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 <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define reps(i,s,n) for(int (i)=(s);(i)<(n);(i)++)
#define repr(i,n) for(int (i)=(n)-1;(i)>=0;(i)--)
#define all(v) (v).begin(),(v).end()
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
#define pb push_back
#define MOD 1000000007
#define inf 1000000000
#define llinf 1000000000000000
#define MAX 1000000


typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int,int> pint;
typedef pair<ll,ll> pll;
typedef vector<pint> vpint;

map<int,int> factor(int m){
    map<int,int> vec;
    int end=pow(m,0.5)+2;
    reps(i,2,end){
        while(m%i==0){
            m/=i;
            vec[i]++;
        }
    }
    return vec;
}

ll fac[MAX],inv[MAX],finv[MAX];

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; 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;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


int main(){
    int n,m;
    cin>>n>>m;
    if(m==1){
        cout<<1<<endl;
        return 0;
    }
    ll ans=1;
    COMinit();
    map<int,int> vec=factor(m);
    for(auto iter:vec){
        ans*=COM(iter.second+n-1,n-1);
        ans%=MOD;
    }
    cout<<ans<<endl;
    
}