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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
using namespace std;
typedef long long  ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
using namespace std;
typedef long long  ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)

ll MOD = 1e9 + 7;
ll f[1000001] = {};
void fact() {
	f[0] = 1;
	rep(i, 1, 1000001)f[i] = f[i - 1] * i %MOD;
}

ll p(ll e, ll n) {
	if (n == 0)return 1;
	if (n == 1)return e;

	ll t = p(e, n / 2);
	if (n & 1) {
		return ((t*t % MOD)*e) % MOD;
	}
	else {
		return t * t % MOD;
	}
}

ll nCr(ll n, ll r) {
	return ((f[n] * p(f[r], MOD - 2) % MOD) * p(f[n - r], MOD - 2) % MOD);
}

map<ll, ll> prime;

int main() {
	fact();
	ll n, m; cin >> n >> m;

	ll t = m;
	for (int i = 2; t != 1; i++) {
		while (!(t % i)) {
			prime[i]++;
			t /= i;
		}
		//while (t != 1) {
		//	if (t % i == 0) {
		//		prime[i]++;
		//		t /= i;
		//	}
		//	else {
		//		break;
		//	}
		//}
	}

	ll ans = 1;
	//rep(i, 0, 100001) {
	//	if (prime[i]) {
	//		ans = ans * nCr(prime[i] + n - 1, prime[i]) % MOD;
	//	}
	//}
	for (map<ll, ll>::iterator it = prime.begin(); it != prime.end(); it++) {
		ans = (ans * nCr(n - 1 + it->second, it->second) % MOD);
	}
	cout << ans << endl;

	return 0;
}