Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.

Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.

Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.

Input

The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.

Output

The only line should contain your answer modulo 10^9+7.

Examples

Input


2


Output


1

Input


3


Output


4

Input


6


Output


120

Note

Consider the second example: these are the permutations of length 3:

  * [1,2,3], f(p)=1. 
  * [1,3,2], f(p)=1. 
  * [2,1,3], f(p)=2. 
  * [2,3,1], f(p)=2. 
  * [3,1,2], f(p)=2. 
  * [3,2,1], f(p)=2. 



The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 50, mod = 1e9 + 7;
int n, k = 0, ans = 1, J[N], I[N], a[60];
int P(int n, int m) { return 1ll * J[n] * I[n - m] % mod; }
int power(int x, int y) {
  int z = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) z = 1ll * z * x % mod;
  return z;
}
int main() {
  scanf("%d", &n);
  J[0] = 1;
  for (int i = 1; i <= n; i++) J[i] = 1ll * J[i - 1] * i % mod;
  I[n] = power(J[n], mod - 2);
  for (int i = n; i; i--) I[i - 1] = 1ll * I[i] * i % mod;
  while ((1 << k + 1) <= n) k++;
  for (int i = 0; i <= k + 1; i++) a[i] = 1 << i;
  for (int i = k, d; ~i; i--) {
    int x = n / a[i] - n / a[i + 1];
    ans = 1ll * ans * x % mod * P(n - n / a[i + 1] - 1, x - 1) % mod;
  }
  if (n > 1 && (1 << k - 1) * 3 <= n) {
    for (int j = 1; j <= k; j++) {
      for (int i = 0; i < j; i++) a[i] = 1 << i;
      a[j] = a[j - 1] * 3;
      for (int i = j + 1; i <= k + 1; i++) a[i] = a[i - 1] << 1;
      int dat = 1;
      for (int i = k, d; ~i; i--) {
        int x = n / a[i] - n / a[i + 1];
        dat = 1ll * dat * x % mod * P(n - n / a[i + 1] - 1, x - 1) % mod;
      }
      (ans += dat) %= mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}
