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#include <bits/stdc++.h>
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using namespace std;
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#define int long long
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#define endl '\n'
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const int MOD = 1e9 + 7;
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// 60 ms
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// 欧拉筛
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const int N = 1e6 + 10;
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int primes[N], cnt; // primes[]存储所有素数
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bool st[N]; // st[x]存储x是否被筛掉
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void get_primes(int n) {
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memset(st, 0, sizeof st);
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cnt = 0;
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for (int i = 2; i <= n; i++) {
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if (!st[i]) primes[cnt++] = i;
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for (int j = 0; primes[j] * i <= n; j++) {
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st[primes[j] * i] = true;
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if (i % primes[j] == 0) break;
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}
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}
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}
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signed main() {
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int n;
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cin >> n;
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// 步骤1:筛质数
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get_primes(n);
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// 步骤2：阶乘质因子分解
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int res = 1;
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for (int i = 0; i < cnt; i++) {
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int p = primes[i];
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int s = 0;
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for (int j = n; j; j /= p) s += j / p; // 应除尽除
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// 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
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res = res * (2 * s + 1) % MOD;
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}
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cout << res << endl;
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}
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