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 inf = 1e9;
const long long INF = 1e18;
const int M = 1e9 + 7;
const int maxn = 1e6;
int d[maxn + 1][22][2];
long long f[30][2];
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  int x = 0, y = 0, tx = 0;
  for (int i = 0; i < 20; ++i) {
    if (n / (1 << i) > 0) {
      x = i;
    }
    if (n / 3 / (1 << i) > 0) {
      tx = i;
    }
  }
  if (n >= 3 && tx == x - 1) {
    ++y;
    d[1][x - 1][1] = 1;
  }
  d[1][x][0] = 1;
  for (int i = 0; i <= x; ++i) {
    for (int j = 0; j <= y; ++j) {
      int tot = 1 << i;
      if (j) {
        tot *= 3;
      }
      f[i][j] = n / tot;
    }
  }
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j <= x; ++j) {
      for (int k = 0; k <= y; ++k) {
        if (f[j][k] > i)
          d[i + 1][j][k] = (d[i + 1][j][k] + d[i][j][k] * (f[j][k] - i)) % M;
        if (j)
          d[i + 1][j - 1][k] =
              (d[i + 1][j - 1][k] + d[i][j][k] * (f[j - 1][k] - f[j][k])) % M;
        if (k)
          d[i + 1][j][k - 1] =
              (d[i + 1][j][k - 1] + d[i][j][k] * (f[j][k - 1] - f[j][k])) % M;
      }
    }
  }
  cout << d[n][0][0] << endl;
  return 0;
}
