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>
const int S = 1000003, mod = 1000000007;
int n, f[S][21][2];
int main() {
  scanf("%d", &n);
  int up = 1;
  for (up = 1; (1 << up) <= n; ++up)
    ;
  --up;
  f[1][up][0] = 1;
  if ((1 << up - 1) * 3 <= n) f[1][up - 1][1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= up; j++) {
      f[i][j][0] =
          (f[i][j][0] +
           1ll * f[i - 1][j][1] * (n / (1 << j) - n / ((1 << j) * 3)) % mod) %
          mod;
      for (int k = 0; k <= 1; ++k) {
        int o = n / ((1 << j) * (k ? 3 : 1));
        f[i][j][k] =
            (f[i][j][k] + 1ll * f[i - 1][j][k] * (o - (i - 1)) % mod) % mod;
        if (j < up)
          f[i][j][k] = (f[i][j][k] + 1ll * f[i - 1][j + 1][k] *
                                         (n / ((1 << j) * (k ? 3 : 1)) -
                                          n / ((1 << j + 1) * (k ? 3 : 1))) %
                                         mod) %
                       mod;
      }
    }
  printf("%d\n", f[n][0][0]);
  return 0;
}
