Given two positive integers **N** and **M**, the task is to find the number of ways to place **M** distinct objects in partitions of even indexed boxes which are numbered **[1, N]** sequentially, and every **i**^{th} Box has **i **distinct partitions. Since the answer can be very large, print modulo 1000000007.

**Examples:**

Input:N = 2, M = 1Output:2Explanation:Since, N = 2. There is only one even indexed box i.e box 2, having 2 partitions. Therefore, there are two positions to place an object. Therefore, number of ways = 2.

Input:N = 5, M = 2Output:32

**Approach: **Follow the steps below to solve the problem:

**M**objects are to be placed in even indexed box’s partitions. Let**S**be the total even indexed box’s partitions in**N boxes.**- The number of partitions is equal to the summation of all even numbers up to
**N.**Therefore, Sum,**S = X * (X + 1),**where**X = floor(N / 2).** - Each object can occupy one of
**S**different positions. Therefore, the total number of ways**= S*S*S..(M times) =****S**^{M}.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the` `// above Aapproach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `const` `int` `MOD = 1000000007;` `// Iterative Function to calculate` `// (x^y)%p in O(log y)` `int` `power(` `int` `x, unsigned ` `int` `y, ` `int` `p = MOD)` `{` ` ` `// Initialize Result` ` ` `int` `res = 1;` ` ` `// Update x if x >= MOD` ` ` `// to avoid multiplication overflow` ` ` `x = x % p;` ` ` `while` `(y > 0) {` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y & 1)` ` ` `res = (res * 1LL * x) % p;` ` ` `// multiplied by long long int,` ` ` `// to avoid overflow` ` ` `// becauuse res * x <= 1e18, which is` ` ` `// out of bounds for integer` ` ` `// n must be even now` ` ` `// y = y/2` ` ` `y = y >> 1;` ` ` `// Change x to x^2` ` ` `x = (x * 1LL * x) % p;` ` ` `}` ` ` `return` `res;` `}` `// Utility function to find` `// the Total Number of Ways` `void` `totalWays(` `int` `N, ` `int` `M)` `{` ` ` `// Number of Even Indexed` ` ` `// Boxes` ` ` `int` `X = N / 2;` ` ` `// Number of paritions of` ` ` `// Even Indexed Boxes` ` ` `int` `S = (X * 1LL * (X + 1)) % MOD;` ` ` `// Number of ways to distribute` ` ` `// objects` ` ` `cout << power(S, M, MOD) << ` `"\n"` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `// N = number of boxes` ` ` `// M = number of distinct objects` ` ` `int` `N = 5, M = 2;` ` ` `// Function call to` ` ` `// get Total Number of Ways` ` ` `totalWays(N, M);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the` `// above Aapproach` `import` `java.io.*;` `class` `GFG` `{` ` ` `public` `static` `int` `MOD = ` `1000000007` `;` `// Iterative Function to calculate` `// (x^y)%p in O(log y)` `static` `int` `power(` `int` `x, ` `int` `y, ` `int` `p)` `{ ` ` ` `p = MOD;` ` ` ` ` `// Initialize Result` ` ` `int` `res = ` `1` `;` ` ` `// Update x if x >= MOD` ` ` `// to avoid multiplication overflow` ` ` `x = x % p;` ` ` `while` `(y > ` `0` `)` ` ` `{` ` ` `// If y is odd, multiply x with result` ` ` `if` `((y & ` `1` `) != ` `0` `)` ` ` `res = (res * x) % p;` ` ` `// multiplied by long long int,` ` ` `// to avoid overflow` ` ` `// becauuse res * x <= 1e18, which is` ` ` `// out of bounds for integer` ` ` `// n must be even now` ` ` `// y = y/2` ` ` `y = y >> ` `1` `;` ` ` `// Change x to x^2` ` ` `x = (x * x) % p;` ` ` `}` ` ` `return` `res;` `}` `// Utility function to find` `// the Total Number of Ways` `static` `void` `totalWays(` `int` `N, ` `int` `M)` `{` ` ` ` ` `// Number of Even Indexed` ` ` `// Boxes` ` ` `int` `X = N / ` `2` `;` ` ` `// Number of paritions of` ` ` `// Even Indexed Boxes` ` ` `int` `S = (X * (X + ` `1` `)) % MOD;` ` ` `// Number of ways to distribute` ` ` `// objects` ` ` `System.out.println(power(S, M, MOD));` `}` `// Driver Code` `public` `static` `void` `main (String[] args)` `{` ` ` ` ` `// N = number of boxes` ` ` `// M = number of distinct objects` ` ` `int` `N = ` `5` `, M = ` `2` `;` ` ` `// Function call to` ` ` `// get Total Number of Ways` ` ` `totalWays(N, M);` `}` `}` `// This code is contributed by Dharanendra L V` |

## Python3

`# Python3 implementation of the` `# above Aapproach` `MOD ` `=` `1000000007` `# Iterative Function to calculate` `# (x^y)%p in O(log y)` `def` `power(x, y, p ` `=` `MOD):` ` ` ` ` `# Initialize Result` ` ` `res ` `=` `1` ` ` `# Update x if x >= MOD` ` ` `# to avoid multiplication overflow` ` ` `x ` `=` `x ` `%` `p` ` ` `while` `(y > ` `0` `):` ` ` `# If y is odd, multiply x with result` ` ` `if` `(y & ` `1` `):` ` ` `res ` `=` `(res ` `*` `x) ` `%` `p` ` ` `# multiplied by long long int,` ` ` `# to avoid overflow` ` ` `# becauuse res * x <= 1e18, which is` ` ` `# out of bounds for integer` ` ` `# n must be even now` ` ` `# y = y/2` ` ` `y ` `=` `y >> ` `1` ` ` `# Change x to x^2` ` ` `x ` `=` `(x ` `*` `x) ` `%` `p` ` ` `return` `res` `# Utility function to find` `# the Total Number of Ways` `def` `totalWays(N, M):` ` ` ` ` `# Number of Even Indexed` ` ` `# Boxes` ` ` `X ` `=` `N ` `/` `/` `2` ` ` `# Number of paritions of` ` ` `# Even Indexed Boxes` ` ` `S ` `=` `(X ` `*` `(X ` `+` `1` `)) ` `%` `MOD` ` ` `# Number of ways to distribute` ` ` `# objects` ` ` `print` `(power(S, M, MOD))` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# N = number of boxes` ` ` `# M = number of distinct objects` ` ` `N, M ` `=` `5` `, ` `2` ` ` `# Function call to` ` ` `# get Total Number of Ways` ` ` `totalWays(N, M)` `# This code is contributed by mohit kumar 29.` |

## C#

`// C# implementation of the` `// above Aapproach` `using` `System;` `public` `class` `GFG{` `public` `static` `int` `MOD = 1000000007;` `// Iterative Function to calculate` `// (x^y)%p in O(log y)` `static` `int` `power(` `int` `x, ` `int` `y, ` `int` `p)` `{` ` ` ` ` `p = MOD;` ` ` ` ` `// Initialize Result` ` ` `int` `res = 1;` ` ` `// Update x if x >= MOD` ` ` `// to avoid multiplication overflow` ` ` `x = x % p;` ` ` `while` `(y > 0) {` ` ` `// If y is odd, multiply x with result` ` ` `if` `((y & 1) != 0)` ` ` `res = (res * x) % p;` ` ` `// multiplied by long long int,` ` ` `// to avoid overflow` ` ` `// becauuse res * x <= 1e18, which is` ` ` `// out of bounds for integer` ` ` `// n must be even now` ` ` `// y = y/2` ` ` `y = y >> 1;` ` ` `// Change x to x^2` ` ` `x = (x * x) % p;` ` ` `}` ` ` `return` `res;` `}` `// Utility function to find` `// the Total Number of Ways` `static` `void` `totalWays(` `int` `N, ` `int` `M)` `{` ` ` ` ` `// Number of Even Indexed` ` ` `// Boxes` ` ` `int` `X = N / 2;` ` ` `// Number of paritions of` ` ` `// Even Indexed Boxes` ` ` `int` `S = (X * (X + 1)) % MOD;` ` ` `// Number of ways to distribute` ` ` `// objects` ` ` `Console.WriteLine(power(S, M, MOD));` `}` `// Driver Code` `static` `public` `void` `Main ()` `{` ` ` `// N = number of boxes` ` ` `// M = number of distinct objects` ` ` `int` `N = 5, M = 2;` ` ` `// Function call to` ` ` `// get Total Number of Ways` ` ` `totalWays(N, M);` `}` `}` `// This code is contributed by Dharanendra L V` |

## Javascript

`<script>` `// Javascript implementation of the` `// above Aapproach` `var` `MOD = 1000000007;` `// Iterative Function to calculate` `// (x^y)%p in O(log y)` `function` `power(x, y, p = MOD)` `{` ` ` `// Initialize Result` ` ` `var` `res = 1;` ` ` `// Update x if x >= MOD` ` ` `// to avoid multiplication overflow` ` ` `x = x % p;` ` ` `while` `(y > 0) {` ` ` `// If y is odd, multiply x with result` ` ` `if` `(y & 1)` ` ` `res = (res * 1 * x) % p;` ` ` `// multiplied by long long int,` ` ` `// to avoid overflow` ` ` `// becauuse res * x <= 1e18, which is` ` ` `// out of bounds for integer` ` ` `// n must be even now` ` ` `// y = y/2` ` ` `y = y >> 1;` ` ` `// Change x to x^2` ` ` `x = (x * 1 * x) % p;` ` ` `}` ` ` `return` `res;` `}` `// Utility function to find` `// the Total Number of Ways` `function` `totalWays(N, M)` `{` ` ` `// Number of Even Indexed` ` ` `// Boxes` ` ` `var` `X = parseInt(N / 2);` ` ` `// Number of paritions of` ` ` `// Even Indexed Boxes` ` ` `var` `S = (X * 1 * (X + 1)) % MOD;` ` ` `// Number of ways to distribute` ` ` `// objects` ` ` `document.write( power(S, M, MOD) << ` `"<br>"` `);` `}` `// Driver Code` `// N = number of boxes` `// M = number of distinct objects` `var` `N = 5, M = 2;` `// Function call to` `// get Total Number of Ways` `totalWays(N, M);` `</script>` |

**Output:**

36

**Time Complexity:** O(log M)**Auxiliary Space:** O(1)

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