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: Time Limit Exceeded)

import java.util.*;
public class Main {
    static long MOD = 1000000007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int res = m;
        long ans = 1;
        for (int i = 2; i * i <= res; i++) {
            int cnt = 0;
            while (res % i == 0) {
                cnt++;
                res /= i;
            }
            ans *= calcComb(cnt + n - 1, n - 1);
            ans %= MOD;
        }
        if (res != 1) {
            ans *= calcComb(n, 1);
            ans %= MOD;
        }
        System.out.println(ans);
    }
    public static long calcComb(int n, int k) {
        long ansMul = 1;
        long ansDiv = 1;
        for (int i = 0; i < k; i++) {
            ansMul *= (long) (n - i);
            ansDiv *= (long) (i + 1);
            ansMul %= MOD;
            ansDiv %= MOD;
        }
        long ans = ansMul * pow(ansDiv, MOD - 2);
        ans %= MOD;
        return ans;
    }
    public static long pow(long x, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = ans * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return ans;
    }
}