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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define ll long long 
#define FI first 
#define SE second
#define pb push_back 
#define eb emplace_back
#define mod 1000000007
#define all(c) (c).begin(),(c).end()
#define LB  lower_bound
#define UB  upper_bound
#define max3(a,b,c) max(c,max(a,b))
#define min3(a,b,c) min(c,min(a,b))
#define mems(s, n) memset(s, n, sizeof(s))
#define NINF -1e18
#define INF 1e18
#define int ll int
#define OOK order_of_key  //no of elements less than
#define FBO find_by_order //iterator pointing kth element;indexing starts from 0
#define CK3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define CK4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl

typedef pair<int,int> PII;
typedef pair<pair<int,int>,int> PPII;
typedef pair<int,pair<int,int>> PIPI;
typedef map<int,int>MII;
typedef vector<int> VI;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

int power(int a, int b)
{int x=1,y=a;while(b>0){if(b%2){x =(x*y)%mod;}y =(y*y)%mod;b/=2;}return x%mod;}

int modInverse(int a, int m) {return power(a, m-2);} //gcd(a,m)=1 
int fact[1000005];int fact_inv[1000005];
int ncr(int n,int r){
    if(n<r){return 0;}
    if(n==r||r==0){return 1;}
    return (((fact[n]*fact_inv[n-r])%mod)*fact_inv[r])%mod;
    //return x; 
}
signed main()
{
ios::sync_with_stdio(false); cin.tie(0);
int T=1,T1=0;//cin>>T;
while(T1++<T)
{//cout<<"Case #"<<T1<<": ";
    int n,m;
    cin>>n>>m;
    fact[0]=1;
    for(int i=1;i<1000005;i++){
        fact[i]=(fact[i-1]*i)%mod;
        fact_inv[i]=power(fact[i],mod-2);
    }


    VI re;int anss=1;
    for(int i=2;i<=sqrt(m);i++)
    {   int flag=0;int ans=0;
        while(m%i==0)
        { flag=1;
            m/=i;ans++;
        }
        if(flag==1)re.pb(ans);
    }
    if(m!=1)re.pb(1);
    
    for(auto it:re)
    {
    	anss*=ncr(n+it-1,n-1);
    	anss%=mod;
    }
    cout<<anss<<endl;
}


return 0;
}