## solution

$n = p_1^{k_1} p_2^{k_2} \dots p_l^{k_l}$と素因数分解できるとき約数の個数$d(n) = (1 + k_1)(1 + k_2)\dots (1 + k_l)$なので、$N!$を構成する$1,2,\dots,N$のそれぞれを素因数分解して指数部を足し合わせる。

## implementation

#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#define repeat_from(i,m,n) for (int i = (m); (i) < int(n); ++(i))
using ll = long long;
using namespace std;

vector<int> sieve_of_eratosthenes(int n) { // enumerate primes in [2,n] with O(n log log n)
vector<bool> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i*i <= n; ++i)
if (is_prime[i])
for (int k = i+i; k <= n; k += i)
is_prime[k] = false;
vector<int> primes;
for (int i = 2; i <= n; ++i)
if (is_prime[i])
primes.push_back(i);
return primes;
}
map<ll,int> prime_factrorize(ll n, vector<int> const & primes) {
map<ll,int> result;
for (int p : primes) {
if (n < p *(ll) p) break;
while (n % p == 0) {
result[p] += 1;
n /= p;
}
}
if (n != 1) result[n] += 1;
return result;
}

const int mod = 1e9+7;
int main() {
int n; cin >> n;
vector<int> primes = sieve_of_eratosthenes(sqrt(n) + 3);
map<ll,int> factors;
repeat_from (i,1,n+1) {
for (auto it : prime_factrorize(i, primes)) {
int p, cnt; tie(p, cnt) = it;
factors[p] += cnt;
}
}
ll ans = 1;
for (auto it : factors) {
int cnt = it.second;
ans = ans * (1 + cnt) % mod;
}
cout << ans << endl;
return 0;
}