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: Wrong Answer)

#include <stdio.h>
#define MOD 1000000007
#define MAX 1001

int main() {
  int primes[MAX];
  int primeNum = 0;
  int factor[MAX][MAX];
  for (int i = 0; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      factor[i][j] = 0;
    }
  }
  int N;
  scanf("%d", &N);

  for (int i = 2; i <= N; i++) {
    int isPrime = 1;
    int minPrimeIndex = primeNum;
    for (int j = 0; j < primeNum; j++) {
      if (i % primes[j] == 0) {
        isPrime = 0;
        minPrimeIndex = j;
        break;
      }
    }
    if (isPrime) {
      primes[primeNum] = i;
      factor[i][primeNum] = 1;
      primeNum++;
    } else {
      for (int j = 0; j < primeNum; j++) {
        factor[i][j] = factor[i / primes[minPrimeIndex]][j];
      }
      factor[i][minPrimeIndex]++;
    }
  }

  long result = 1;
  for (int i = 0; i < primeNum; i++) {
    int numOfI = 0;
    for (int j = 2; j <= N; j++) {
      numOfI += factor[j][i];
    }
    printf("[%d]:%d\n", primes[i], numOfI);
    result *= (numOfI + 1);
    result %= MOD;
  }

  printf("%ld\n", result);
  return 0;
}