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>

#define fi first
#define se second
#define rep(i, n) for (ll i=0;i<(n); i++)
#define rrep(i, n) for (ll (i) = n-1;(i) >= 0;--(i))
#define ALL(v) (v).begin(), (v).end()
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pi;

//const ll M=1000000007ll;
const ll MOD=1000000007ll;
//const ll MOD=2019ll;
const ll MAX=100000001ll;
//const ll M = 1000000000+7;

const ll mod = 1000000007ll;
//const int mod = 2019;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
   return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(ll n){
  if(n==0) return 1;
  mint ans = f(n/2);
  ans *= ans;
  if(n%2==1) ans*= 2;
  return ans;
}

mint g(ll n, ll x){
  if(x==0) return 1;
  mint ans = g(n, x/2);
  ans *= ans;
  if(x%2==1) ans*= n;
  return ans;  
}

mint choose(ll n, ll k){
  mint x=1, y=1;
  rep(i,k){
    x *= (n-i);
    y *= (i+1);
  }
  //cout <<"#" <<x.x<<" "<<y.x<<endl;
  return x / y;
}

struct BIT {
  private:
   vector<ll> bit;
   ll N;
 
  public:
   BIT(ll size) {
     N = size;
     bit.resize(N + 1);
   }
 
   // 一点更新です
   void add(ll a, ll w) {
     for (ll x = a; x <= N; x += x & -x) bit[x] += w;
   }
 
   // 1~Nまでの和を求める。
   int sum(ll a) {
     ll ret = 0;
     for (ll x = a; x > 0; x -= x & -x) ret += bit[x];
     return ret;
   }
 };
 
ll fff(ll n, vector<ll>& v, ll i, ll j, ll rl){
  ll can = 0;
  ll can_i = 0;

  cout << v[0]<<endl;
  
  // right
  if(rl == 0){
    rep(k, n){
      if(v[k]*(j-k)>can){
	can = v[k]*(j-k);
	can_i = k;
      }
      //cout << can<<endl;
    }
    v[can_i] = 0;
    return can;
  }
  // left
  else{
    rep(k, n){
      if(v[k]*(k-i)>can){
	can = v[k]*(k-i);
	can_i = k;
      }
      //cout << can<<endl;

    }
    v[can_i] = 0;
    return can;
  }
  
}

ll ff(ll x){
  return x * (x+1) / 2;
}


void calc_prime(vector<ll> &p){
  for(ll i=2;i<=p.size()+1;i++){
    if(p[i]<0){
      for(ll j=2;j<=p.size()+1;j++){
	if(i%j==0){
	  p[i] = j;
	  break;
	} 
      }
      if(p[i]<0) p[i]=i;
    }
  }
}

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

  mint ans = 1;
  ll tmp = m;
  map<ll, ll> ma;
  
  for(ll i=2;(i)*(i)<=m;i++){
    while(tmp%i==0){
      ma[i]++;
      tmp /= i;
    }
  }
  
  //ma[m]++;
  ll cnt = 0;
  for(auto x: ma){
    cnt++;
    ans *= choose(x.se+n-1, x.se);
    //    cout<<ans.x<<endl;
  }
  if(cnt==0) ans *= n;
  cout << ans.x <<endl;
}