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: Wrong Answer)

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
#include <complex> // complex<double> a(1.2 , 2.3);// real(): 1.2, imag()2.3
using namespace std;

#define MOD 1000000007
#define ll long long
#define ld long double
#define FOR(i,a,b) for(ll i=(ll)a;i<(ll)b;i++)
#define rep(i,n) FOR(i,0,n)
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rmsame(a) sort(all(a)),a.erase(unique(all(a)), a.end())
#define rmvector(a,b) rep(i,a.size())rep(j,b.size())if(a[i]==b[j]){a.erase(a.begin()+i);i--;break;}
#define pq_pair_tB priority_queue <pair<ll,ll>, vector<pair<ll,ll> > , greater<pair<ll,ll> > >
#define pq_pair_ts priority_queue <pair<ll,ll> > //第二成分の大きさが関係ない
template<typename X> bool exist(vector<X> vec, X item){return find(all(vec), item)!=vec.end();}
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll N, K;
vector<ll> P ;

void cal(){
    rep(i, 100000){
        if(i <= 1) continue;
        ll cnt = 0;

        while(K % i == 0 ){
            cnt ++;
            K /= i;
        }
        if(cnt != 0) P.pb(cnt);
    }
    if(K != 1) P.pb(1);
}


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N >> K;
    cal();

    cerr << "素因数分解まで" << endl;
    rep(i, P.size()){
        cout << P[i] << endl;
    }


	//cout << fixed << setprecision(16) << ans << endl;
	return 0;
}