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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.awt.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    static long mod=1000000007;
    public static void main(String[] args){
        Scanner sc = new Scanner(in);
        int size=10000001;
        long[] fact=new long[size];
        long[] inv=new long[size];
        fact[0]=1;
        inv[0]=1;
        for (int i = 1; i < size; i++) {
            fact[i]=(fact[i-1]*i)%mod;
            inv[i]=power(fact[i],(int)mod-2)%mod;
        }
        //comb=fact[n]*inv[k]%mod*inv[n-k]%mod
        int n=sc.nextInt();
        int m=sc.nextInt();
        if(m==1){
            out.println(1);
        }else{
            HashMap<Integer,Integer> map=new HashMap<>();
            int Stop=(int)(sqrt(m)+1);
            for (int i = 2; i <= Stop; i++) {
                while(m%i==0){
                    m/=i;
                    map.put(i,map.getOrDefault(i, 0)+1);
                }
            }
            if(map.isEmpty())map.put(m,1);
            long ans=1;
            for(int k:map.keySet()){
                ans=(ans*fact[n-1+map.get(k)]*inv[n-1]%mod*inv[map.get(k)]%mod)%mod;
            }
            out.println(ans);
        }
    }
    static long power(long x,int n){
        long ans=1;
        while (n>0) {
            if((n&1)==1){
                ans=ans*x%mod;
            }
            x=x*x%mod;
            n>>=1;
        }
        return ans;
    }
}