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: Runtime Error)


#include <stdio.h>
#include <string.h>

long combination(int n, int r)
{
	if (r == 0 || r == n){
		return (1);
	}else if (r == 1){
		return (n);
	}
	return ((combination(n - 1, r - 1) + combination(n - 1, r))%1000000007);
}

int main(void)
{
	//変数の宣言
	int n,m;
	int ans=0;
	int i,j,k;
	int insucount[4200];
	int insutotal;
	
	
	//素数生成用
	int sosu[40000];
	int sosulen;
	int warukazu;
	
	
	
	//データの読み込み
	scanf("%d %d",&n,&m);
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	//40000までの素数列の作製
	
	for(i=0;i<40000;i++){
		sosu[i]=i+2;
	}
	sosu[40001]=-1;
	
	i=0;
	
	while(sosu[i]!=-1){
//		printf("%d番目の素数は%dです\n",i+1,sosu[i]);
		warukazu=sosu[i];
		j=i+1;
		k=i+1;
		while(sosu[j]!=-1){
			if(sosu[j]%sosu[i]!=0){
				sosu[k]=sosu[j];
				k++;
			}
			j++;
		}
		sosu[k+1]=-1;
		i++;
	}
	
	//4200番目まで使う
	
//	printf("素数生成完了\n");
	
	j=0;
	insucount[0]=0;
	
	for(i=0;i<4200;i++){
//		printf("%d番目の素数について探索中\n",i+1);
		if(m%sosu[i]==0){
			while(m%sosu[i]==0){
				insucount[j]++;
				m=m/sosu[i];
			}
			insucount[j+1]==0;
			j++;
		}
	}
	
	insutotal=j;
	
//	for(i=0;i<insutotal;i++){
//		printf("%d番目の因数は%d個",i+1,insucount[i]);
//	}
	
	ans=1;
//	printf("n=%d\n",n);
	
	for(i=0;i<insutotal;i++){
	//		printf("insucount[%d]=%d\n",i,insucount[i]);
		ans=ans*combination(n+insucount[i]-1,insucount[i]);
		ans=ans%1000000007;
	}
	
	
	
//	printf("計算部分終了\n");
	
	//出力
	
		printf("%d",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
}