Problem:
Score : 400 points

Problem Statement

You are given positive integers N and M.

How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.

Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

2 6

Sample Output 1

4

Four sequences satisfy the condition: \{a_1, a_2\} = \{1, 6\}, \{2, 3\}, \{3, 2\} and \{6, 1\}.

Sample Input 2

3 12

Sample Output 2

18

Sample Input 3

100000 1000000000

Sample Output 3

957870001

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    final int MOD = 1000000007;
    int N = scanner.nextInt();
    int M = scanner.nextInt();

    List<Integer> factorials = new ArrayList<>();
    int k = M;
    for (int i = 2; i <= M && k > 1; i++) {
      int count = 0;
      while (k % i == 0) {
        count++;
        k /= i;
      }
      if (count > 0) factorials.add(count);
    }

    long count = 1;
    for(long f : factorials) {
      count = (count * choose(f + N - 1, f, MOD)) % MOD;
    }
    System.out.println(count);
  }

  public static long choose(long n, long k, long mod) {
    if (k > n - k) return choose(n, n - k, mod);
    long res = 1;
    for (int i = 0; i < k; i++) {
      res = (res * (n - i)) % mod;
    }
    for (int i = 1; i <= k; i++) {
      res = (res * inverse(i, mod)) % mod;
    }
    return res % mod;
  }

  public static long inverse(long a, long mod) {
    return modpow(a, mod - 2, mod);
  }

  public static long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
      if (n % 2 == 1) res = (res * a) % mod;
      a = (a * a) % mod;
      n >>= 1;
    }
    return res % mod;
  }
}