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: Time Limit Exceeded)

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <time.h>
typedef long long int ll;
#define mod 1000000007

/*
int combination(ll n,ll r){
	const ll N = 500;
	ll C[N][N];
	memset(C, 0, sizeof(C));
	C[0][0]=1;
	for(ll n=1; n<N; n++){
		C[n][0]=C[n-1][0];
		for(ll r=1; r<N; r++){
			C[n][r] = C[n-1][r-1] + C[n-1][r];
		}
	}
	return C[n][r];
}
*/

//二分累乗法



//逆元
int inverse(ll a){
	ll res = 1;
	ll n = mod-2;
	while(n>0) {
		if(n&1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1; 
	}
	return res;
}

int combination(ll n,ll r){
	const ll N = 500;
	ll x=1,y=1;
	for(int i=1;i<=r;i++){
		x = x * (n-i+1) % mod;
		y = y * i % mod;
	}
	return x * inverse(y) % mod;
}

int main(void){
	ll n,m;
	scanf("%lld %lld",&n,&m);
	ll x,i=0,ans=1;
	for(x=2;m>1;x++){
		i=0;
		while((m%x)==0){
			m /=x;
			i++;
		}
		ans=ans*combination(n-1+i,i)%mod;
	}
	printf("%lld",ans);
	return 0;
}