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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

// Define
using ll = int64_t;
using ul = uint64_t;
using ld = long double;
const ll MOD = 1e9 + 7;
const ll INF = LONG_MAX;
const ul MAX = ULONG_MAX;
const char nl = '\n';

#define elif else if
#define def inline auto
#define run __attribute__((constructor)) def _

// Debug
#define debug(x) cerr << (x) << " (L:" << __LINE__ << ")" << '\n'

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i < _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i > _##i; --i)
#define rep(i, n) inc(i, 0, n)
#define be(c) begin(c), end(c)

// Stream
#define fout(n) cout << fixed << setprecision(n)
#define fasten cin.tie(0), ios::sync_with_stdio(0)

// Speed
run() { fasten, fout(10); }
#pragma GCC optimize("-O3")
#pragma GCC target("avx")

map<ll, int> prime_factor(ll n) {
    ll a = 2;
    map<ll, int> ans;
    while (n >= a * a) {
        if (n % a == 0) {
            ++ans[a];
            n /= a;
        } else {
            a++;
        }
    }
    if (n != 1) ans[n] = 1;
    return ans;
}

signed main() {
    int n;
    cin >> n;
    auto mp = prime_factor(n);
    ll ans = 1;
    for(auto& e : mp){
        ans *= e.second + 1;
        ans %= MOD;
    }
    cout << ans << nl;
}