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 + 5, L = 21, M = 1e9 + 7;
int n;
int f[N][L][2];
int wk(int x, int y) {
  if (y) return n / 3 / (1 << x);
  return n / (1 << x);
}
int main() {
  int l = 1;
  scanf("%d", &n);
  while ((1 << l) <= n) l++;
  l--;
  f[1][l][0] = 1;
  if ((1 << (l - 1)) * 3 <= n) f[1][l - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int x = 0; x <= l; x++) {
      for (int y = 0; y <= 1; y++) {
        f[i][x][y] = 1ll * f[i - 1][x][y] * (wk(x, y) - (i - 1)) % M;
        if (x < l)
          f[i][x][y] = (f[i][x][y] +
                        1ll * f[i - 1][x + 1][y] * (wk(x, y) - wk(x + 1, y))) %
                       M;
        if (y < 1)
          f[i][x][y] = (f[i][x][y] +
                        1ll * f[i - 1][x][y + 1] * (wk(x, y) - wk(x, y + 1))) %
                       M;
      }
    }
  }
  printf("%d\n", f[n][0][0]);
  return 0;
}
