Given an array **arr[][]** consisting of **Q** queries where every row consists of two numbers **L** and **R** which denotes the range **[L, R]**; the task is to find the sum of the product of proper divisors of all numbers lying in the range [L, R].

**Note:** Since the answer might be very big, perform % with 1000000007 for every query.

**Examples:**

Input:Q = 2, arr[] = { { 4, 6 }, { 8, 10 } }

Output:9 21

Explanation:

Query 1: From 4 to 6

Product of proper divisors of 4 = 1 * 2 = 2

Product of proper divisors of 5 = 1

Product of proper divisors of 6 = 1 * 2 * 3 = 6

Sum of product of proper divisors from 4 to 6 = 2 + 1 + 6 = 9

Query 2: From 8 to 10

Product of proper divisors of 8 = 1 * 2 * 4 = 8

Product of proper divisors of 9 = 1 * 3 = 3

Product of proper divisors of 10 = 1 * 2 * 5 = 10

Sum of product of proper divisors from 8 to 10 = 8 + 3 + 10 = 21

Input:Q = 2, arr[] = { { 10, 20 }, { 12, 16 } }

Output:975 238

**Approach:** Since there can be multiple queries and finding the divisors and product for every query is not feasible, the idea is to precompute and store every element along with its product of proper divisors in an array using a modification of Sieve of Eratosthenes.

Once the product of the proper divisors of all the numbers is stored in an array, the idea is to use the concept of prefix sum array. The sum of the product of proper divisors till that particular index is precomputed and stored in an array **pref[]** so that every query can be answered in constant time.

For each query, the sum of all product of proper divisors for the range **[L, R]** can be found as follows:

sum = pref[R] - pref[L - 1]

Below is the implementation of the above approach:

## CPP

`// C++ implementation to find the sum ` `// of the product of proper divisors of ` `// all the numbers lying in the range [L, R] ` ` ` `#include <bits/stdc++.h> ` `#define ll long long int ` `#define mod 1000000007 ` `using` `namespace` `std; ` ` ` `// Vector to store the product ` `// of the proper divisors of a number ` `vector<ll> ans(100002, 1); ` ` ` `// Variable to store the prefix ` `// sum of the product array ` `long` `long` `pref[100002]; ` ` ` `// Function to precompute the product ` `// of proper divisors of a number at ` `// it's corresponding index ` `void` `preCompute() ` `{ ` ` ` `// Modificatino of sieve to store the ` ` ` `// product of the proper divisors ` ` ` `for` `(` `int` `i = 2; i <= 100000 / 2; i++) { ` ` ` `for` `(` `int` `j = 2 * i; j <= 100000; j += i) { ` ` ` ` ` `// Multiplying the existing value ` ` ` `// with i because i is the ` ` ` `// proper divisor of ans[j] ` ` ` `ans[j] = (ans[j] * i) % mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Loop to store the prefix sum of the ` ` ` `// previously computed product array ` ` ` `for` `(` `int` `i = 1; i < 100002; ++i) { ` ` ` ` ` `// Computing the prefix sum ` ` ` `pref[i] = pref[i - 1] ` ` ` `+ ans[i]; ` ` ` `pref[i] %= mod; ` ` ` `} ` `} ` ` ` `// Function to print the sum ` `// for each query ` `void` `printSum(` `int` `L, ` `int` `R) ` `{ ` ` ` `cout << pref[R] - pref[L - 1] ` ` ` `<< ` `" "` `; ` `} ` ` ` `// Function to print te sum of product ` `// of proper divisors of a number in ` `// [L, R] ` `void` `printSumProper(` `int` `arr[][2], ` `int` `Q) ` `{ ` ` ` ` ` `// Calling the function that ` ` ` `// pre computes ` ` ` `// the sum of product ` ` ` `// of proper divisors ` ` ` `preCompute(); ` ` ` ` ` `// Iterate over all Queries ` ` ` `// to print the sum ` ` ` `for` `(` `int` `i = 0; i < Q; i++) { ` ` ` `printSum(arr[i][0], arr[i][1]); ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `Q = 2; ` ` ` `int` `arr[][2] = { { 10, 20 }, ` ` ` `{ 12, 16 } }; ` ` ` ` ` `printSumProper(arr, Q); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation to find the sum ` `// of the product of proper divisors of ` `// all the numbers lying in the range [L, R] ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `static` `int` `mod = ` `1000000007` `; ` ` ` `// Vector to store the product ` `// of the proper divisors of a number ` `static` `int` `[]ans = ` `new` `int` `[` `100002` `]; ` ` ` `// Variable to store the prefix ` `// sum of the product array ` `static` `int` `[]pref = ` `new` `int` `[` `100002` `]; ` ` ` `// Function to precompute the product ` `// of proper divisors of a number at ` `// it's corresponding index ` `static` `void` `preCompute() ` `{ ` ` ` `// Modificatino of sieve to store the ` ` ` `// product of the proper divisors ` ` ` `Arrays.fill(ans, ` `1` `); ` ` ` `for` `(` `int` `i = ` `2` `; i <= ` `100000` `/ ` `2` `; i++) { ` ` ` `for` `(` `int` `j = ` `2` `* i; j <= ` `100000` `; j += i) { ` ` ` ` ` `// Multiplying the existing value ` ` ` `// with i because i is the ` ` ` `// proper divisor of ans[j] ` ` ` `ans[j] = (ans[j] * i) % mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Loop to store the prefix sum of the ` ` ` `// previously computed product array ` ` ` `for` `(` `int` `i = ` `1` `; i < ` `100002` `; ++i) { ` ` ` ` ` `// Computing the prefix sum ` ` ` `pref[i] = pref[i - ` `1` `] ` ` ` `+ ans[i]; ` ` ` `pref[i] %= mod; ` ` ` `} ` `} ` ` ` `// Function to print the sum ` `// for each query ` `static` `void` `printSum(` `int` `L, ` `int` `R) ` `{ ` ` ` `System.out.print(pref[R] - pref[L - ` `1` `]+` `" "` `); ` `} ` ` ` `// Function to print te sum of product ` `// of proper divisors of a number in ` `// [L, R] ` `static` `void` `printSumProper(` `int` `[][]arr, ` `int` `Q) ` `{ ` ` ` ` ` `// Calling the function that ` ` ` `// pre computes ` ` ` `// the sum of product ` ` ` `// of proper divisors ` ` ` `preCompute(); ` ` ` ` ` `// Iterate over all Queries ` ` ` `// to print the sum ` ` ` `for` `(` `int` `i = ` `0` `; i < Q; i++) { ` ` ` `printSum(arr[i][` `0` `], arr[i][` `1` `]); ` ` ` `} ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `Q = ` `2` `; ` ` ` `int` `[][] arr = {{` `10` `, ` `20` `}, ` ` ` `{ ` `12` `, ` `16` `} }; ` ` ` ` ` `printSumProper(arr, Q); ` `} ` `} ` ` ` `// This code is contributed by Surendra_Gangwar ` |

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## Python3

`# Python3 implementation to find the sum ` `# of the product of proper divisors of ` `# all the numbers lying in the range [L, R] ` ` ` `mod ` `=` `1000000007` ` ` `# Vector to store the product ` `# of the proper divisors of a number ` `ans ` `=` `[` `1` `]` `*` `(` `100002` `) ` ` ` `# Variable to store the prefix ` `# sum of the product array ` `pref ` `=` `[` `0` `]` `*` `100002` ` ` `# Function to precompute the product ` `# of proper divisors of a number at ` `# it's corresponding index ` `def` `preCompute(): ` ` ` ` ` `# Modificatino of sieve to store the ` ` ` `# product of the proper divisors ` ` ` `for` `i ` `in` `range` `(` `2` `,` `100000` `/` `/` `2` `+` `1` `): ` ` ` `for` `j ` `in` `range` `(` `2` `*` `i,` `100000` `+` `1` `,i): ` ` ` ` ` `# Multiplying the existing value ` ` ` `# with i because i is the ` ` ` `# proper divisor of ans[j] ` ` ` `ans[j] ` `=` `(ans[j] ` `*` `i) ` `%` `mod ` ` ` ` ` `# Loop to store the prefix sum of the ` ` ` `# previously computed product array ` ` ` `for` `i ` `in` `range` `(` `1` `,` `100002` `): ` ` ` ` ` `# Computing the prefix sum ` ` ` `pref[i] ` `=` `pref[i ` `-` `1` `]` `+` `ans[i] ` ` ` `pref[i] ` `%` `=` `mod ` ` ` `# Function to prthe sum ` `# for each query ` `def` `printSum(L, R): ` ` ` ` ` `print` `(pref[R] ` `-` `pref[L ` `-` `1` `],end` `=` `" "` `) ` ` ` `# Function to prte sum of product ` `# of proper divisors of a number in ` `# [L, R] ` `def` `printSumProper(arr, Q): ` ` ` ` ` `# Calling the function that ` ` ` `# pre computes ` ` ` `# the sum of product ` ` ` `# of proper divisors ` ` ` `preCompute() ` ` ` ` ` `# Iterate over all Queries ` ` ` `# to prthe sum ` ` ` `for` `i ` `in` `range` `(Q): ` ` ` `printSum(arr[i][` `0` `], arr[i][` `1` `]) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `Q ` `=` `2` ` ` `arr` `=` `[ [ ` `10` `, ` `20` `], ` ` ` `[ ` `12` `, ` `16` `] ] ` ` ` ` ` `printSumProper(arr, Q) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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**Output:**

975 238

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