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 <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define all(v) v.begin(), v.end()
using in = int64_t;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
const double PI=3.14159265358979323846;
//const in MOD = 1000000007;
const in INF=1<<30;
#define vec2(a,y,x) vector<vector<int>> a(y,vector<int>(x))
#define vec1(a,n) vector<int> a(n);rep(i,n)cin>>a[i]
using P=pair<int,int>;

const int MAX = 510000;
const int MOD = 1000000007;

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

// テーブルを作る前処理
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
in 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 nhr(int n,int r){return COM(n+r-1ll,r);}

int main(){
  COMinit();
  int n,m;
  cin>>n>>m;
  vector<int> soi(0);
  int x=2;
  while(m>1){
    if(m%x==0){ m/=x; soi.push_back(x);}
    else x ++;
  }
  sort(all(soi));
  auto y=soi.begin();
  vector<int> sp(0);
  while(y<soi.end()){
    sp.push_back(upper_bound(all(soi),*y)-lower_bound(all(soi),*y));
    y=upper_bound(all(soi),*y);
  }
  in ans=1ll;
  rep(i,sp.size()){
  	ans *=nhr(n,(sp[i]));
    ans%=MOD;
  }
  cout<<ans<<endl;
}