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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>
#include <string>
#include <numeric>
#include <algorithm>
#include <utility>


using std::cout;
using std::cin;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;


ll gcd(ll a, ll b) {
    while (b != 0) {
        auto tmp = a;
        a = b;
        b = tmp % a;
    }
    return a;
}

ll lcm(ll a, ll b) {
    return (a * b) / gcd(a, b);
}

int isPrime(int a) {
    for (int i = 2; i * i <= a; ++i) {
        if (a % i == 0) {
            return 0;
        }
    }
    return 1;
}

int cnt(int a, int p) {
    int d = p;
    int res = 0;
    while (d <= a) {
        res += a / d;
        d *= p;
    }
    return res;
}



void oneCase () {
    int n;
    cin >> n;
    vector<int> d;
    for (int i = 2; i <= n; ++i) {
        if (isPrime(i) == 0) {
            continue;
        }
        d.push_back(1 + cnt(n, i));
    }
    ll res = 1;
    ll mod = 1e9 + 7;
    for (auto v : d) {
        res = (res * v) % mod;
    }
    cout << res << "\n";
}





int main() {
    int t = 1;
//    cin >> t;
    while (t > 0) {
        oneCase();
        --t;
    }

    return 0;
}