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>
#include <string.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define chmax(a,b) ((a)=(a)>(b)?(a):(b))
#define chmin(a,b) ((a)=(a)<(b)?(a):(b))
#define abs(p) ((p)>=(0)?(p):(-(p)))
#define MOD 1000000007
ll power(ll a,ll b){ll r=1;rep(i,b){r*=a;}return r;}
#define swap(a,b) do{ll w=(a);(a)=(b);(b)=w;}while(0)
#define swapd(a,b) do{double w=(a);(a)=(b);(b)=w}while(0)

//your code here!
#define N_MAX 1001
ll prime[N_MAX], cnt;
int main(void){
    ll N, ans=1, n;
    scanf("%lld", &N);
    for(ll i=2; i<=N; i++) {
        n = i;
        for(ll j=2; j*j<=n; j++) {
            while(n%j==0) {
                n /= j;
                prime[j]++;
            }
        }
        if(n>1) prime[n]++;
    }
    rep(i, N_MAX) if(prime[i]) {
        ans *= 1+prime[i];
        ans %= MOD;
    }
    printf("%lld\n", ans);
    return 0;
}