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 MOD = 1e9 + 7;
const int MAX_N = 1e6 + 5;
int f[MAX_N][21][3];
int g(int x, int y, int n) { return n / ((1 << x) * (y ? 3 : 1)); }
int main() {
  int n;
  cin >> n;
  int pow = 0;
  int nn = n;
  while (nn > 1) {
    nn /= 2;
    ++pow;
  }
  f[1][pow][0] = 1;
  if ((1 << (pow - 1)) * 3 <= n) {
    f[1][pow - 1][1] = 1;
  }
  for (int i = 1; i <= n - 1; ++i) {
    for (int x = 0; x <= pow; ++x) {
      for (int y = 0; y <= 1; ++y) {
        f[i + 1][x][y] =
            (f[i + 1][x][y] + (f[i][x][y] * 1ll * (g(x, y, n) - i)) % MOD) %
            MOD;
        f[i + 1][x - 1][y] =
            (f[i + 1][x - 1][y] +
             (f[i][x][y] * 1ll * (g(x - 1, y, n) - g(x, y, n))) % MOD) %
            MOD;
        if (y) {
          f[i + 1][x][y - 1] =
              (f[i + 1][x][y - 1] +
               (f[i][x][y] * 1ll * (g(x, y - 1, n) - g(x, y, n))) % MOD) %
              MOD;
        }
      }
    }
  }
  cout << f[n][0][0] << "\n";
  return 0;
}
