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>
long long md = 7 + pow(10, 9);
int dp[1000005][21][2];
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long i, j, k, n;
  int x;
  cin >> n;
  x = (int)log2(n);
  dp[1][x][0] = 1;
  if (n >= 3 * pow(2, x - 1)) dp[1][x - 1][1] = 1;
  for (i = 1; i < n; ++i) {
    for (j = 0; j < x + 1; ++j) {
      for (k = 0; k < 2; ++k) {
        dp[i + 1][j][k] =
            (dp[i + 1][j][k] +
             (1ll * dp[i][j][k] * (n / ((1 << j) * (2 * (k) + 1)) - i))) %
            md;
        if (j)
          dp[i + 1][j - 1][k] =
              (dp[i + 1][j - 1][k] + (1ll * dp[i][j][k] *
                                      (n / ((1 << j - 1) * (2 * (k) + 1)) -
                                       n / ((1 << j) * (2 * (k) + 1))))) %
              md;
        if (k)
          dp[i + 1][j][k - 1] =
              (dp[i + 1][j][k - 1] + (1ll * dp[i][j][k] *
                                      (n / ((1 << j) * (2 * (k - 1) + 1)) -
                                       n / ((1 << j) * (2 * (k) + 1))))) %
              md;
      }
    }
  }
  cout << dp[n][0][0];
  return 0;
}
