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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

int main(){
  int N, kaijo = 1, count = 0, remainder;
  scanf("%d", &N);

  for(int i = 0; i < N; i++){
    kaijo *= (N - i);//N!を求める
  }

  for (int i = 1; i < kaijo + 1; i++){
    if(kaijo % i == 0){
      count += 1;//N！の正の約数の個数
  }}

  remainder = count % ( (int)pow(10, 9) + 7 );
  printf("%d\n", remainder );
  return 0;
}