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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int MOD = 1e9 + 7;
// 60 ms
// 欧拉筛
const int N = 1e6 + 10;
int primes[N], cnt; // primes[]存储所有素数
bool st[N]; // st[x]存储x是否被筛掉
void get_primes(int n) {
memset(st, 0, sizeof st);
cnt = 0;
for (int i = 2; i <= n; i++) {
if (!st[i]) primes[cnt++] = i;
for (int j = 0; primes[j] * i <= n; j++) {
st[primes[j] * i] = true;
if (i % primes[j] == 0) break;
}
}
}
signed main() {
int n;
cin >> n;
// 步骤1:筛质数
get_primes(n);
// 步骤2阶乘质因子分解
int res = 1;
for (int i = 0; i < cnt; i++) {
int p = primes[i];
int s = 0;
for (int j = n; j; j /= p) s += j / p; // 应除尽除
// 步骤3约数个数公式.因为通过公式推导我们知道不是n!,而是(n!)^2,所以,是(2*s+1)做为累乘因子
res = res * (2 * s + 1) % MOD;
}
cout << res << endl;
}