Problem:
Score : 300 points

Problem Statement

You are given an integer N.
Find the number of the positive divisors of N!, modulo 10^9+7.

Constraints

1≤N≤10^3

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the positive divisors of N!, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

4

There are four divisors of 3! =6: 1, 2, 3 and 6. Thus, the output should be 4.

Sample Input 2

6

Sample Output 2

30

Sample Input 3

1000

Sample Output 3

972926972

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define M 1000000007
int factor(int n,int i){
    int ans=0,temp=i;
    //printf("i=%d",i);
    while(n/i!=0){
        //printf(" n/i=%d",n/i);
        ans+=n/i;
        i*=temp;
    }
    //printf(" ans=%d\n",ans);
    return ans;
}
int main(void){
    int n;
    scanf("%d",&n);
    int i,j,flag;
    long long int ans=1;
    for(i=2;i<=n;i++){
        flag=1;
        for(j=2;j<=sqrt(i);j++){
            if(i%j==0){
                flag=0;
                break;}
            }
            //素数であれば
            if(flag==1){
                ans=(ans*(factor(n,i)+1))%M;
            }
        }
    printf("%lld\n",ans);
}