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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

long long mod=1000000007;
long long modp(long long d,long long p){
    if (p==0){
        return 1 ;
    } else if (p==1){
        return d%mod ;
    } else if (p%2==0){
        return ((modp(d,p/2)%mod)*(modp(d,p/2))%mod)%mod;
    } else {
        return ((modp(d,p/2)%mod)*(modp(d,p/2))%mod)*d%mod;
    }
}

long long ncr(long long a,long long b){
    long long combinationans=1;
    for (long long i=0; i<b; i++){
        combinationans=(combinationans*(a-i)*modp((i+1),mod-2))%mod;
    }
    return (combinationans);
}

int main(void){
    long long n,m,p,ans=1;
    scanf("%lld%lld",&n,&m);
    p=m;
    for (int i=2; i<=p; i++){
        long long count=0;
        while(p%i==0){
            p/=i;
            count++;
        }
        if (count!=0){
            ans=(ans*ncr(count+n-1,count))%mod;
        }
    }
    printf("%lld\n",ans);
    return 0;
}