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<math.h>
int num_of_factors(int n,int k){
  int ans=0;
  int m=k;
  while(n/k!=0){
    ans+=n/k;
    k*=m;
  }
  return ans;
}
     
int main(){
  int n,i,j,flag,a;
  long ans=1;
  scanf("%d",&n);
  for(i=2;i<=n;i++){
    flag=1;
    for(j=2;j<=(int)sqrt(i);j++){
      if(i%j==0) flag=0;
    }
    if(flag==1) {
      a=num_of_factors(n,i);
      ans=(ans*(a+1))%1000000007;
    }
  }
  printf("%ld\n",ans);
  return 0;
}