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 mod = 1000000007, maxn = 1000005, maxl = 25;
int n, max2;
int f[maxn][maxl][2];
inline void add(int a, int b, int c, int x, int y, int z, int k) {
  f[a][b][c] = (f[a][b][c] + 1ll * f[x][y][z] * k % mod) % mod;
}
int main() {
  scanf("%d", &n);
  for (max2 = 1; (1 << (max2 + 1)) <= n; max2++)
    ;
  f[1][max2][0] = 1;
  if ((1 << (max2 - 1)) * 3 <= n) f[1][max2 - 1][1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= max2; j++) {
      int k = 1 << j;
      add(i, j, 0, i - 1, j, 0, (n / k) - (i - 1));
      if (j < max2) add(i, j, 0, i - 1, j + 1, 0, n / k - n / (k * 2));
      add(i, j, 0, i - 1, j, 1, n / k - n / (k * 3));
      add(i, j, 1, i - 1, j, 1, n / (k * 3) - (i - 1));
      if (j < max2) add(i, j, 1, i - 1, j + 1, 1, n / (k * 3) - n / (k * 6));
    }
  printf("%d\n", f[n][0][0]);
  return 0;
}
