# Count Inversions in an array | Set 1 (Using Merge Sort)

Inversion Count for an array indicates – how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j

Example:

```Input: arr[] = {8, 4, 2, 1}
Output: 6

Explanation: Given array has six inversions:
(8,4), (4,2),(8,2), (8,1), (4,1), (2,1).

Input: arr[] = {3, 1, 2}
Output: 2

Explanation: Given array has two inversions:
(3, 1), (3, 2)
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

METHOD 1 (Simple)

• Approach :Traverse through the array and for every index find the number of smaller elements on its right side of the array. This can be done using a nested loop. Sum up the counts for all index in the array and print the sum.
• Algorithm :
1. Traverse through the array from start to end
2. For every element find the count of elements smaller than the current number upto that index using another loop.
3. Sum up the count of inversion for every index.
4. Print the count of inversions.
• Implementation:

## C++

 `// C++ program to Count Inversions ` `// in an array ` `#include ` `using` `namespace` `std; ` ` `  `int` `getInvCount(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `inv_count = 0; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``if` `(arr[i] > arr[j]) ` `                ``inv_count++; ` ` `  `    ``return` `inv_count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 20, 6, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << ``" Number of inversions are "` `         ``<< getInvCount(arr, n); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## C

 `// C program to Count ` `// Inversions in an array ` `#include ` `int` `getInvCount(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `inv_count = 0; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``if` `(arr[i] > arr[j]) ` `                ``inv_count++; ` ` `  `    ``return` `inv_count; ` `} ` ` `  `/* Driver program to test above functions */` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 20, 6, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``printf``(``" Number of inversions are %d \n"``, getInvCount(arr, n)); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count ` `// inversions in an array ` `class` `Test { ` `    ``static` `int` `arr[] = ``new` `int``[] { ``1``, ``20``, ``6``, ``4``, ``5` `}; ` ` `  `    ``static` `int` `getInvCount(``int` `n) ` `    ``{ ` `        ``int` `inv_count = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) ` `                ``if` `(arr[i] > arr[j]) ` `                    ``inv_count++; ` ` `  `        ``return` `inv_count; ` `    ``} ` ` `  `    ``// Driver method to test the above function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``System.out.println(``"Number of inversions are "` `                           ``+ getInvCount(arr.length)); ` `    ``} ` `} `

## Python3

 `# Python3 program to count  ` `# inversions in an array ` ` `  `def` `getInvCount(arr, n): ` ` `  `    ``inv_count ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``if` `(arr[i] > arr[j]): ` `                ``inv_count ``+``=` `1` ` `  `    ``return` `inv_count ` ` `  `# Driver Code ` `arr ``=` `[``1``, ``20``, ``6``, ``4``, ``5``] ` `n ``=` `len``(arr) ` `print``(``"Number of inversions are"``, ` `              ``getInvCount(arr, n)) ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// C# program to count inversions ` `// in an array ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG { ` ` `  `    ``static` `int``[] arr = ``new` `int``[] { 1, 20, 6, 4, 5 }; ` ` `  `    ``static` `int` `getInvCount(``int` `n) ` `    ``{ ` `        ``int` `inv_count = 0; ` ` `  `        ``for` `(``int` `i = 0; i < n - 1; i++) ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `                ``if` `(arr[i] > arr[j]) ` `                    ``inv_count++; ` ` `  `        ``return` `inv_count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.WriteLine(``"Number of "` `                          ``+ ``"inversions are "` `                          ``+ getInvCount(arr.Length)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` ``\$arr``[``\$j``]) ` `                ``\$inv_count``++; ` ` `  `    ``return` `\$inv_count``; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(1, 20, 6, 4, 5 ); ` `\$n` `= sizeof(``\$arr``); ` `echo` `"Number of inversions are "``,  ` `           ``getInvCount(``\$arr``, ``\$n``); ` ` `  `// This code is contributed by ita_c ` `?> `

Output:

`Number of inversions are 5`

• Complexity Analysis:
• Time Complexity: O(n^2), Two nested loops are needed to traverse the array from start to end so the Time complexity is O(n^2)
• Space Compelxity:O(1), No extra space is required.

METHOD 2(Enhance Merge Sort)

• Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2), what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get a number of inversions, that needs to be added a number of inversions in the left subarray, right subarray and merge(). How to get number of inversions in merge()?
In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j] The complete picture: • Algorithm:
1. The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
2. Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for first half and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
3. Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, number of inversion in the second half and the number of inversions by merging the two.
4. The base case of recursion is when there is only one element in the given half.
• Implementation:

## C++

 `// C++ program to Count ` `// Inversions in an array ` `// using Merge Sort ` `#include ` `using` `namespace` `std; ` ` `  `int` `_mergeSort(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `right); ` `int` `merge(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `mid, ``int` `right); ` ` `  `/* This function sorts the input array and returns the  ` `number of inversions in the array */` `int` `mergeSort(``int` `arr[], ``int` `array_size) ` `{ ` `    ``int` `temp[array_size]; ` `    ``return` `_mergeSort(arr, temp, 0, array_size - 1); ` `} ` ` `  `/* An auxiliary recursive function that sorts the input array and  ` `returns the number of inversions in the array. */` `int` `_mergeSort(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `right) ` `{ ` `    ``int` `mid, inv_count = 0; ` `    ``if` `(right > left) { ` `        ``/* Divide the array into two parts and  ` `        ``call _mergeSortAndCountInv()  ` `        ``for each of the parts */` `        ``mid = (right + left) / 2; ` ` `  `        ``/* Inversion count will be sum of  ` `        ``inversions in left-part, right-part  ` `        ``and number of inversions in merging */` `        ``inv_count += _mergeSort(arr, temp, left, mid); ` `        ``inv_count += _mergeSort(arr, temp, mid + 1, right); ` ` `  `        ``/*Merge the two parts*/` `        ``inv_count += merge(arr, temp, left, mid + 1, right); ` `    ``} ` `    ``return` `inv_count; ` `} ` ` `  `/* This funt merges two sorted arrays  ` `and returns inversion count in the arrays.*/` `int` `merge(``int` `arr[], ``int` `temp[], ``int` `left, ` `          ``int` `mid, ``int` `right) ` `{ ` `    ``int` `i, j, k; ` `    ``int` `inv_count = 0; ` ` `  `    ``i = left; ``/* i is index for left subarray*/` `    ``j = mid; ``/* j is index for right subarray*/` `    ``k = left; ``/* k is index for resultant merged subarray*/` `    ``while` `((i <= mid - 1) && (j <= right)) { ` `        ``if` `(arr[i] <= arr[j]) { ` `            ``temp[k++] = arr[i++]; ` `        ``} ` `        ``else` `{ ` `            ``temp[k++] = arr[j++]; ` ` `  `            ``/* this is tricky -- see above  ` `            ``explanation/diagram for merge()*/` `            ``inv_count = inv_count + (mid - i); ` `        ``} ` `    ``} ` ` `  `    ``/* Copy the remaining elements of left subarray  ` `(if there are any) to temp*/` `    ``while` `(i <= mid - 1) ` `        ``temp[k++] = arr[i++]; ` ` `  `    ``/* Copy the remaining elements of right subarray  ` `(if there are any) to temp*/` `    ``while` `(j <= right) ` `        ``temp[k++] = arr[j++]; ` ` `  `    ``/*Copy back the merged elements to original array*/` `    ``for` `(i = left; i <= right; i++) ` `        ``arr[i] = temp[i]; ` ` `  `    ``return` `inv_count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 20, 6, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `ans = mergeSort(arr, n); ` `    ``cout << ``" Number of inversions are "` `<< ans; ` `    ``return` `0; ` `} ` ` `  `// This is code is contributed by rathbhupendra `

## C

 `// C program to Count ` `// Inversions in an array ` `// using Merge Sort ` `#include ` ` `  `int` `_mergeSort(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `right); ` `int` `merge(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `mid, ``int` `right); ` ` `  `/* This function sorts the input array and returns the ` `   ``number of inversions in the array */` `int` `mergeSort(``int` `arr[], ``int` `array_size) ` `{ ` `    ``int``* temp = (``int``*)``malloc``(``sizeof``(``int``) * array_size); ` `    ``return` `_mergeSort(arr, temp, 0, array_size - 1); ` `} ` ` `  `/* An auxiliary recursive function that sorts the input array and ` `  ``returns the number of inversions in the array. */` `int` `_mergeSort(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `right) ` `{ ` `    ``int` `mid, inv_count = 0; ` `    ``if` `(right > left) { ` `        ``/* Divide the array into two parts and call _mergeSortAndCountInv() ` `       ``for each of the parts */` `        ``mid = (right + left) / 2; ` ` `  `        ``/* Inversion count will be the sum of inversions in left-part, right-part ` `      ``and number of inversions in merging */` `        ``inv_count += _mergeSort(arr, temp, left, mid); ` `        ``inv_count += _mergeSort(arr, temp, mid + 1, right); ` ` `  `        ``/*Merge the two parts*/` `        ``inv_count += merge(arr, temp, left, mid + 1, right); ` `    ``} ` `    ``return` `inv_count; ` `} ` ` `  `/* This funt merges two sorted arrays and returns inversion count in ` `   ``the arrays.*/` `int` `merge(``int` `arr[], ``int` `temp[], ``int` `left, ``int` `mid, ``int` `right) ` `{ ` `    ``int` `i, j, k; ` `    ``int` `inv_count = 0; ` ` `  `    ``i = left; ``/* i is index for left subarray*/` `    ``j = mid; ``/* j is index for right subarray*/` `    ``k = left; ``/* k is index for resultant merged subarray*/` `    ``while` `((i <= mid - 1) && (j <= right)) { ` `        ``if` `(arr[i] <= arr[j]) { ` `            ``temp[k++] = arr[i++]; ` `        ``} ` `        ``else` `{ ` `            ``temp[k++] = arr[j++]; ` ` `  `            ``/*this is tricky -- see above explanation/diagram for merge()*/` `            ``inv_count = inv_count + (mid - i); ` `        ``} ` `    ``} ` ` `  `    ``/* Copy the remaining elements of left subarray ` `   ``(if there are any) to temp*/` `    ``while` `(i <= mid - 1) ` `        ``temp[k++] = arr[i++]; ` ` `  `    ``/* Copy the remaining elements of right subarray ` `   ``(if there are any) to temp*/` `    ``while` `(j <= right) ` `        ``temp[k++] = arr[j++]; ` ` `  `    ``/*Copy back the merged elements to original array*/` `    ``for` `(i = left; i <= right; i++) ` `        ``arr[i] = temp[i]; ` ` `  `    ``return` `inv_count; ` `} ` ` `  `/* Driver program to test above functions */` `int` `main(``int` `argv, ``char``** args) ` `{ ` `    ``int` `arr[] = { 1, 20, 6, 4, 5 }; ` `    ``printf``(``" Number of inversions are %d \n"``, mergeSort(arr, 5)); ` `    ``getchar``(); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function to count the number of inversions ` `    ``// during the merge process ` `    ``private` `static` `int` `mergeAndCount(``int``[] arr, ``int` `l, ``int` `m, ``int` `r) ` `    ``{ ` ` `  `        ``// Left subarray ` `        ``int``[] left = Arrays.copyOfRange(arr, l, m + ``1``); ` ` `  `        ``// Right subarray ` `        ``int``[] right = Arrays.copyOfRange(arr, m + ``1``, r + ``1``); ` ` `  `        ``int` `i = ``0``, j = ``0``, k = l, swaps = ``0``; ` ` `  `        ``while` `(i < left.length && j < right.length) { ` `            ``if` `(left[i] <= right[j]) ` `                ``arr[k++] = left[i++]; ` `            ``else` `{ ` `                ``arr[k++] = right[j++]; ` `                ``swaps += (m + ``1``) - (l + i); ` `            ``} ` `        ``} ` ` `  `        ``// Fill from the rest of the left subarray ` `        ``while` `(i < left.length) ` `            ``arr[k++] = left[i++]; ` ` `  `        ``// Fill from the rest of the right subarray ` `        ``while` `(j < right.length) ` `            ``arr[k++] = right[j++]; ` ` `  `        ``return` `swaps; ` `    ``} ` ` `  `    ``// Merge sort function ` `    ``private` `static` `int` `mergeSortAndCount(``int``[] arr, ``int` `l, ``int` `r) ` `    ``{ ` ` `  `        ``// Keeps track of the inversion count at a ` `        ``// particular node of the recursion tree ` `        ``int` `count = ``0``; ` ` `  `        ``if` `(l < r) { ` `            ``int` `m = (l + r) / ``2``; ` ` `  `            ``// Total inversion count = left subarray count ` `            ``// + right subarray count + merge count ` ` `  `            ``// Left subarray count ` `            ``count += mergeSortAndCount(arr, l, m); ` ` `  `            ``// Right subarray count ` `            ``count += mergeSortAndCount(arr, m + ``1``, r); ` ` `  `            ``// Merge count ` `            ``count += mergeAndCount(arr, l, m, r); ` `        ``} ` ` `  `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = { ``1``, ``20``, ``6``, ``4``, ``5` `}; ` ` `  `        ``System.out.println(mergeSortAndCount(arr, ``0``, arr.length - ``1``)); ` `    ``} ` `} ` ` `  `// This code is contributed by Pradip Basak `

## Python3

 `# Python 3 program to count inversions in an array ` ` `  `# Function to Use Inversion Count ` `def` `mergeSort(arr, n): ` `    ``# A temp_arr is created to store ` `    ``# sorted array in merge function ` `    ``temp_arr ``=` `[``0``]``*``n ` `    ``return` `_mergeSort(arr, temp_arr, ``0``, n``-``1``) ` ` `  `# This Function will use MergeSort to count inversions ` ` `  `def` `_mergeSort(arr, temp_arr, left, right): ` ` `  `    ``# A variable inv_count is used to store ` `    ``# inversion counts in each recursive call ` ` `  `    ``inv_count ``=` `0` ` `  `    ``# We will make a recursive call if and only if ` `    ``# we have more than one elements ` ` `  `    ``if` `left < right: ` ` `  `        ``# mid is calculated to divide the array into two subarrays ` `        ``# Floor division is must in case of python ` ` `  `        ``mid ``=` `(left ``+` `right)``/``/``2` ` `  `        ``# It will calculate inversion counts in the left subarray ` ` `  `        ``inv_count ``+``=` `_mergeSort(arr, temp_arr, left, mid) ` ` `  `        ``# It will calculate inversion counts in right subarray ` ` `  `        ``inv_count ``+``=` `_mergeSort(arr, temp_arr, mid ``+` `1``, right) ` ` `  `        ``# It will merge two subarrays in a sorted subarray ` ` `  `        ``inv_count ``+``=` `merge(arr, temp_arr, left, mid, right) ` `    ``return` `inv_count ` ` `  `# This function will merge two subarrays in a single sorted subarray ` `def` `merge(arr, temp_arr, left, mid, right): ` `    ``i ``=` `left     ``# Starting index of left subarray ` `    ``j ``=` `mid ``+` `1` `# Starting index of right subarray ` `    ``k ``=` `left     ``# Starting index of to be sorted subarray ` `    ``inv_count ``=` `0` ` `  `    ``# Conditions are checked to make sure that i and j don't exceed their ` `    ``# subarray limits. ` ` `  `    ``while` `i <``=` `mid ``and` `j <``=` `right: ` ` `  `        ``# There will be no inversion if arr[i] <= arr[j] ` ` `  `        ``if` `arr[i] <``=` `arr[j]: ` `            ``temp_arr[k] ``=` `arr[i] ` `            ``k ``+``=` `1` `            ``i ``+``=` `1` `        ``else``: ` `            ``# Inversion will occur. ` `            ``temp_arr[k] ``=` `arr[j] ` `            ``inv_count ``+``=` `(mid``-``i ``+` `1``) ` `            ``k ``+``=` `1` `            ``j ``+``=` `1` ` `  `    ``# Copy the remaining elements of left subarray into temporary array ` `    ``while` `i <``=` `mid: ` `        ``temp_arr[k] ``=` `arr[i] ` `        ``k ``+``=` `1` `        ``i ``+``=` `1` ` `  `    ``# Copy the remaining elements of right subarray into temporary array ` `    ``while` `j <``=` `right: ` `        ``temp_arr[k] ``=` `arr[j] ` `        ``k ``+``=` `1` `        ``j ``+``=` `1` ` `  `    ``# Copy the sorted subarray into Original array ` `    ``for` `loop_var ``in` `range``(left, right ``+` `1``): ` `        ``arr[loop_var] ``=` `temp_arr[loop_var] ` `         `  `    ``return` `inv_count ` ` `  `# Driver Code ` `# Given array is ` `arr ``=` `[``1``, ``20``, ``6``, ``4``, ``5``] ` `n ``=` `len``(arr) ` `result ``=` `mergeSort(arr, n) ` `print``(``"Number of inversions are"``, result) ` ` `  `# This code is contributed by ankush_953 `

## C#

 `// C# implementation of counting the ` `// inversion using merge sort ` ` `  `using` `System; ` `public` `class` `Test { ` ` `  `    ``/* This method sorts the input array and returns the ` `       ``number of inversions in the array */` `    ``static` `int` `mergeSort(``int``[] arr, ``int` `array_size) ` `    ``{ ` `        ``int``[] temp = ``new` `int``[array_size]; ` `        ``return` `_mergeSort(arr, temp, 0, array_size - 1); ` `    ``} ` ` `  `    ``/* An auxiliary recursive method that sorts the input array and ` `      ``returns the number of inversions in the array. */` `    ``static` `int` `_mergeSort(``int``[] arr, ``int``[] temp, ``int` `left, ``int` `right) ` `    ``{ ` `        ``int` `mid, inv_count = 0; ` `        ``if` `(right > left) { ` `            ``/* Divide the array into two parts and call _mergeSortAndCountInv() ` `           ``for each of the parts */` `            ``mid = (right + left) / 2; ` ` `  `            ``/* Inversion count will be the sum of inversions in left-part, right-part ` `          ``and number of inversions in merging */` `            ``inv_count += _mergeSort(arr, temp, left, mid); ` `            ``inv_count += _mergeSort(arr, temp, mid + 1, right); ` ` `  `            ``/*Merge the two parts*/` `            ``inv_count += merge(arr, temp, left, mid + 1, right); ` `        ``} ` `        ``return` `inv_count; ` `    ``} ` ` `  `    ``/* This method merges two sorted arrays and returns inversion count in ` `       ``the arrays.*/` `    ``static` `int` `merge(``int``[] arr, ``int``[] temp, ``int` `left, ``int` `mid, ``int` `right) ` `    ``{ ` `        ``int` `i, j, k; ` `        ``int` `inv_count = 0; ` ` `  `        ``i = left; ``/* i is index for left subarray*/` `        ``j = mid; ``/* j is index for right subarray*/` `        ``k = left; ``/* k is index for resultant merged subarray*/` `        ``while` `((i <= mid - 1) && (j <= right)) { ` `            ``if` `(arr[i] <= arr[j]) { ` `                ``temp[k++] = arr[i++]; ` `            ``} ` `            ``else` `{ ` `                ``temp[k++] = arr[j++]; ` ` `  `                ``/*this is tricky -- see above explanation/diagram for merge()*/` `                ``inv_count = inv_count + (mid - i); ` `            ``} ` `        ``} ` ` `  `        ``/* Copy the remaining elements of left subarray ` `       ``(if there are any) to temp*/` `        ``while` `(i <= mid - 1) ` `            ``temp[k++] = arr[i++]; ` ` `  `        ``/* Copy the remaining elements of right subarray ` `       ``(if there are any) to temp*/` `        ``while` `(j <= right) ` `            ``temp[k++] = arr[j++]; ` ` `  `        ``/*Copy back the merged elements to original array*/` `        ``for` `(i = left; i <= right; i++) ` `            ``arr[i] = temp[i]; ` ` `  `        ``return` `inv_count; ` `    ``} ` ` `  `    ``// Driver method to test the above function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = ``new` `int``[] { 1, 20, 6, 4, 5 }; ` `        ``Console.Write(``"Number of inversions are "` `+ mergeSort(arr, 5)); ` `    ``} ` `} ` `// This code is contributed by Rajput-Ji `

Output:

`Number of inversions are 5`
• Complexity Analysis:
• Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level one full array traversal is needed and there are log n levels so the time complexity is O(n log n).
• Space Compelxity: O(n), Temporary array.

Note that above code modifies (or sorts) the input array. If we want to count only inversions then we need to create a copy of original array and call mergeSort() on copy.

Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.

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