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

struct Benri { Benri() { std::cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(12);}} benri;

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vll = vector<long long>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using ull = unsigned long long;

template <typename T> using PQ = priority_queue<T>;
template <typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define mp make_pair
#define F first
#define S second

template<class T>bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }

constexpr long long MOD = 1000000007;
//constexpr long long MOD = 998244353;
//constexpr int INF = 1001001001;
constexpr ll INF = 1001001001001001001ll;
constexpr double EPS = 1e-10;
using number = long long;


template< int64_t mod >
struct mInt {
    int64_t x;

    mInt() : x(0LL) {}
    mInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    mInt &operator+=(const mInt &p) { if ((x += p.x) >= mod) x -= mod; return *this; }
    mInt &operator-=(const mInt &p) {if ((x += mod - p.x) >= mod) x -= mod; return *this;}
    mInt &operator*=(const mInt &p) { x = (x * p.x % mod); return *this;}
    mInt &operator/=(const mInt &p) { *this *= p.inverse(); return *this;}
    mInt operator-() const { return mInt(-x); }
    mInt operator+(const mInt &p) const { return mInt(*this) += p; }
    mInt operator-(const mInt &p) const { return mInt(*this) -= p; }
    mInt operator*(const mInt &p) const { return mInt(*this) *= p; }
    mInt operator/(const mInt &p) const { return mInt(*this) /= p; }
    bool operator==(const mInt &p) const { return x == p.x; }
    bool operator!=(const mInt &p) const { return x != p.x; }
    mInt inverse() const {
        int64_t a = x, b = mod, u = 1, v = 0, t;
        while (b > 0) {t = a / b; swap(a -= t * b, b); swap(u -= t * v, v);}
        return mInt(u);
    }
    mInt pow(int64_t n) const {
        mInt ret(1), mul(x);
        while (n > 0) {if (n & 1) ret *= mul; mul *= mul; n >>= 1;}
        return ret;
    }
    friend ostream &operator<<(ostream &os, const mInt &p) {
        return os << p.x;
    }
    friend istream &operator>>(istream &is, mInt &a) {
        int64_t t; is >> t;
        a = mInt< mod >(t);
        return (is);
    }
    static int64_t get_mod() { return mod; }
};
using mint = mInt< MOD >;


vector< bool > prime_table(number n) {
    vector< bool > prime(n + 1, true);
    if (n >= 0) prime[0] = false;
    if (n >= 1) prime[1] = false;
    for (number i = 2; i * i <= n; i++) {
        if (!prime[i]) continue;
        for (number j = i + i; j <= n; j += i) {
            prime[j] = false;
        }
    }
    return prime;
}

int main() {
    int n; cin >> n;
    auto pr = prime_table(n);
    mint ans = 1;
    rep(p, n + 1) {
        if (!pr[p]) continue;
        mint tmp = 1;
        int beki = p;
        while (n >= beki) {
            tmp += (n / beki);
            beki *= p;
        }
        ans *= tmp;
    }

    cout << ans << endl;
}