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 <string.h>

int main(void){
  int N;
  scanf("%d",&N);
  int i,j;
  int sosu[500]={
    2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,
    103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,
    199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,
    313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,
    433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,
    563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,
    673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,
    811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,
    941,947,953,967,971,977,983,991,997
  };
  int rog;
  int box[169] = {0};
  for(i=2;i<=N;i++){
    rog = i;
    loop:
    for(j=0;j<168;j++){
      if(i%sosu[j] != 0){
        continue;
      }else if(i % sosu[j] == 0 && i / sosu[j] == 1){
        box[j]++;
        break;
      }else i = i / sosu[j]; box[j]++; goto loop;
    }
    i = rog;
  }
  long sum = 1;
  for(i=0;i<=168;i++){
    sum *= box[i] + 1;
    sum = sum % 1000000007;
  }

  printf("%ld\n",sum);
}