Given an integer **N**, the task is to find the minimum positive product of first** N – 1** natural numbers, i.e. **[1, (N – 1)]**, by swapping any **i**^{th} bit of any two numbers any number of times.

**Note:** N is always a perfect power of 2. Since the product can be very large, print the answer modulo **10 ^{9} + 7**.

**Examples:**

Input:N = 4Output:6Explanation:

No swapping of bits is required. Therefore, the minimum product is 1*2*3 = 6.

Input:N = 8Output:1512Explanation:

Let the array arr[] stores all the value from 1 to N as {1, 2, 3, 4, 5, 6, 7}

Follow the below steps:

Step 1: In elements 2 = (0010) and 5 = (0101), swap 0th and 1st bit. Therefore, replace 2 with 1 and 5 with 6. arr[] = {1, 1, 3, 4, 6, 6, 7}.

Step 2: In elements 3 = (0011) and 4 = (0100), swap 1th bit. Therefore, replace 3 with 1 and 4 with 6. arr[] = {1, 1, 1, 6, 6, 6, 7}.

Hence, the minimum product = 1*1*1*6*6*6*7 = 1512 % 1e9+7 = 1512.

**Approach:** The idea is to make some observations. For example, if **N **= **8 **and **arr[] = {1, 2, 3, 4, 5, 6, 7}**, observe that for the product to be minimum there must be three sixes i.e., there must be an element having value** (N – 2)** with the frequency of occurrence as **(1 + (N – 4)/2)** and there must be three ones i.e., there must be **(1 + (N – 4)/2) **ones. And at last multiply the current product with **(N – 1)**. Hence, the formula becomes:

Minimum product for any value N = ((N – 1) * (N – 2)^{(N – 4)/2 + 1}) % 1e9 + 7

Follow the below steps to solve the problem:

- Initialize the
**ans**as**1**. - Iterate over the range
**[0, 1 + (N – 4)/2]**. - In each traversal, multiply
**ans**with**N – 2**and update the**ans**to**ans mod 1e9+7**. - After the above steps, print the value of
**ans*(N – 1) mod 1e9+7**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `mod = 1e9 + 7;` `// Function to find the minimum product` `// of 1 to N - 1 after performing the` `// given operations` `void` `minProduct(` `int` `n)` `{` ` ` `// Initialize ans with 1` ` ` `int` `ans = 1;` ` ` `// Multiply ans with N-2` ` ` `// ((N - 4)/2) times` ` ` `for` `(` `int` `i = 1;` ` ` `i <= (n - 4) / 2; i++) {` ` ` `ans = (1LL * ans` ` ` `* (n - 2))` ` ` `% mod;` ` ` `}` ` ` `// Multiply ans with N - 1` ` ` `// and N - 2 once` ` ` `ans = (1LL * ans` ` ` `* (n - 2) * (n - 1))` ` ` `% mod;` ` ` `// Print ans` ` ` `cout << ans << endl;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given Number N` ` ` `int` `N = 8;` ` ` `// Function Call` ` ` `minProduct(N);` ` ` `return` `0;` `}` |

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## Java

`// Java program for the ` `// above approach` `import` `java.util.*;` `class` `GFG{` `static` `int` `mod = (` `int` `)1e9 + ` `7` `;` `// Function to find the ` `// minimum product of 1 ` `// to N - 1 after performing ` `// the given operations` `static` `void` `minProduct(` `int` `n)` `{` ` ` `// Initialize ans with 1` ` ` `int` `ans = ` `1` `;` ` ` `// Multiply ans with N-2` ` ` `// ((N - 4)/2) times` ` ` `for` `(` `int` `i = ` `1` `;` ` ` `i <= (n - ` `4` `) / ` `2` `; i++) ` ` ` `{` ` ` `ans = (` `int` `)(1L * ans * ` ` ` `(n - ` `2` `)) % mod;` ` ` `}` ` ` `// Multiply ans with N - 1` ` ` `// and N - 2 once` ` ` `ans = (` `int` `)(1L * ans * ` ` ` `(n - ` `2` `) * (n - ` `1` `)) % mod;` ` ` `// Print ans` ` ` `System.out.print(ans + ` `"\n"` `);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `// Given Number N` ` ` `int` `N = ` `8` `;` ` ` `// Function Call` ` ` `minProduct(N);` `}` `}` `// This code is contributed by gauravrajput1` |

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## Python3

`# Python3 program for the above approach` `mod ` `=` `1e9` `+` `7` `# Function to find the minimum product` `# of 1 to N - 1 after performing the` `# given operations` `def` `minProduct(n):` ` ` ` ` `# Initialize ans with 1` ` ` `ans ` `=` `1` ` ` `# Multiply ans with N-2` ` ` `# ((N - 4)/2) times` ` ` `for` `i ` `in` `range` `(` `1` `, (n ` `-` `4` `) ` `/` `/` `2` `+` `1` `):` ` ` `ans ` `=` `(ans ` `*` `(n ` `-` `2` `)) ` `%` `mod` ` ` `# Multiply ans with N - 1` ` ` `# and N - 2 once` ` ` `ans ` `=` `(ans ` `*` `(n ` `-` `2` `) ` `*` `(n ` `-` `1` `)) ` `%` `mod` ` ` `# Print ans` ` ` `print` `(` `int` `(ans))` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given number N` ` ` `N ` `=` `8` ` ` `# Function call` ` ` `minProduct(N)` `# This code is contributed by mohit kumar 29` |

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## C#

`// C# program for the ` `// above approach` `using` `System;` `class` `GFG{` `static` `int` `mod = (` `int` `)1e9 + 7;` `// Function to find the ` `// minimum product of 1 ` `// to N - 1 after performing ` `// the given operations` `static` `void` `minProduct(` `int` `n)` `{` ` ` `// Initialize ans with 1` ` ` `int` `ans = 1;` ` ` `// Multiply ans with N-2` ` ` `// ((N - 4)/2) times` ` ` `for` `(` `int` `i = 1;` ` ` `i <= (n - 4) / 2; i++) ` ` ` `{` ` ` `ans = (` `int` `)(1L * ans * ` ` ` `(n - 2)) % mod;` ` ` `}` ` ` `// Multiply ans with N - 1` ` ` `// and N - 2 once` ` ` `ans = (` `int` `)(1L * ans * ` ` ` `(n - 2) * ` ` ` `(n - 1)) % mod;` ` ` `// Print ans` ` ` `Console.Write(ans + ` `"\n"` `);` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `// Given Number N` ` ` `int` `N = 8;` ` ` `// Function Call` ` ` `minProduct(N);` `}` `}` `// This code is contributed by Rajput-Ji` |

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**Output**

1512

**Time Complexity:** O(N) where N is the given integer.**Auxiliary Space:** O(1)

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