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: Runtime Error)

import java.util.*;
import java.awt.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static int MAX=1000000000;
    public static int MOD=1000000007;
    public static void main(String[] args){
        Scanner sc = new Scanner(in);
        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;
        COMinit();
        for(int k:map.keySet()){
            ans=(ans*comb(n-1+map.get(k),n-1))%MOD;
        }
        out.println(ans);
        }
    }
//二項係数
//前処理
static final long[] fac=new long[MAX];
static final long[] finv=new long[MAX];
static final long[] inv=new long[MAX];
static 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
static long comb (int a, int b) {
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return fac[a]*(finv[b]*finv[a-b]%MOD)%MOD;
}
//順列
static long permu(long a,long b){
    long temp=1;
    for (long i =a ; i > a-b; i--) {
        temp*=i;
    }
    return temp;
}
}