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 <fstream>
#include <cmath>  
#include <cstdlib>
#include <ctime>
#include <algorithm>
#include <numeric>
#include <functional>
#include <string> 
#include <vector>
#include <bitset>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>

using namespace std;
using ll = long long;

template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;

#define REP(i,n) for(int i = 0; i < int(n); i++)
#define FOR(i, m, n) for(int i = int(m);i < int(n);i++)
#define ALL(obj) (obj).begin(),(obj).end()

const ll MOD = (ll)1e9 + 7;
const ll HINF = (ll)1e18;
const ll LINF = (ll)1e9;
const long double PI = 3.1415926535897932384626433;

template<class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl;exit(0);}else return;}
template <class T, class U>ostream &operator<<(ostream &o, const map<T, U>&obj) {o << "{"; for (auto &x : obj) o << " {" << x.first << " : " << x.second << "}" << ","; o << " }"; return o;}
template <class T>ostream &operator<<(ostream &o, const set<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const vector<T>&obj) {o << "{"; for (int i = 0; i < (int)obj.size(); ++i)o << (i > 0 ? ", " : "") << obj[i]; o << "}"; return o;}
template <class T, class U>ostream &operator<<(ostream &o, const pair<T, U>&obj) {o << "{" << obj.first << ", " << obj.second << "}"; return o;}
template <template <class tmp>  class T, class U> ostream &operator<<(ostream &o, const T<U> &obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr)o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
void print(void) {cout << endl;}
template <class Head> void print(Head&& head) {cout << head;print();}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head << " ";print(forward<Tail>(tail)...);}
void YN(bool flg) {cout << ((flg) ? "YES" : "NO") << endl;}
void Yn(bool flg) {cout << ((flg) ? "Yes" : "No") << endl;}
void yn(bool flg) {cout << ((flg) ? "yes" : "no") << endl;}

//Combination Mod
class Combination_Mod {
public:
	vector<long long> fac, finv, inv;
	long long MOD;

	Combination_Mod(int N, long long mod) : fac(vector<long long>(N + 1)), finv(vector<long long>(N + 1)), inv(vector<long long>(N + 1)), MOD(mod) {
		fac[0] = fac[1] = finv[0] = finv[1] = inv[1] = 1;
		for (int i = 2; i <= N; ++i) {
			fac[i] = fac[i - 1] * i % MOD;
			inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
			finv[i] = finv[i - 1] * inv[i] % MOD;
		}
	}

	long long num(int n, int k) {
		return ((n < 0 || k < 0 || n < k) ? 0 : fac[n] * (finv[k] * finv[n - k] % MOD) % MOD);
	}
};

//Prime Factorization
vector<long long> PrimeFact(long long N) {
	vector<long long> ret;
	if (N == 1) ret.push_back(1);
	for (long long i = 2, M = N; i*i <= M; ++i) {
		if (N%i == 0) ret.push_back(i);
		while (N%i == 0) N /= i;
	}
	if (N != 1) ret.push_back(N);
	return ret;
}

int main() {
	ll N, M, ans = 1; cin >> N >> M;
	Combination_Mod CM(150000LL, MOD);
	auto p = PrimeFact(M);
	for(auto u:p){
		if (u == 1) continue;
		ll cnt = 0;
		while(M%u==0){
			M /= u;
			cnt++;
		}
		(ans *= CM.num(N - 1 + cnt, cnt))%= MOD;
	}
	cout << ans << endl;

	return 0;
}