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 (GCC 5.4.1) (judge verdict: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); bool s=false; if(c=='-'){s=true;c=getchar_unlocked();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); int s=0; if(c=='-'){s=1;c=getchar_unlocked();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar_unlocked(); while(c==' '||c=='\n') c=getchar_unlocked(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar_unlocked(); } *s='\0'; return len; }

#define P 1000000007

int k[200000];
int rk[200000];

int getk(int a, int x){
	if(x==1) return a;
	int b = getk(a, x/2);
	b = ((long)b)*b%P;
	if((x&1)==0){
		return b;
	}else{
		return ((long)b)*a%P;
	}
}

int comb(int a, int b){
	if(a<b) return 0;
	return ((long)k[a])*rk[b]%P*rk[a-b]%P;
}

void comb_init(int n){
	k[0]=1;
	for(int i=1; i<=n; i++){
		k[i]=k[i-1]*((long)i)%P;
	}	
	rk[n]=getk(k[n], P-2);
	for(int i=n-1; i>=0; i--){
		rk[i]=((long)rk[i+1])*(i+1)%P;
	}
}

int a[25];
int main(void){
	int n=nextint();
	int m=nextint();
	int q=sqrt(m);
	int k=0;
	int max=1;
	for(int i=2; i<=q; i++){
		if(m%i==0){
			while(m%i==0){ m/=i; a[k]++; }
			if(a[k]>max) max=a[k];
			k++;
		}
	}
	if(m>1){
		a[k++]=1;
	}
	comb_init(n+max);
	long ans=1;
	for(int i=0; i<k; i++){
		ans=ans*comb(n+a[i]-1, a[i])%P;
	}
	printf("%ld\n", ans);
}