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;
int dp[1000005][21][2];
long long int f(int n, int x, int y) {
  int z = (1 << x);
  if (y) z *= 3;
  return n / z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int te = 1, MOD = 1e9 + 7;
  while (te--) {
    int n;
    cin >> n;
    int x = log2(n);
    dp[1][x][0] = 1;
    if ((1 << (x - 1)) * 3 <= n) dp[1][x - 1][1] = 1;
    for (int i = 1; i < n; i++) {
      for (int j = x; j >= 0; j--) {
        for (int k = 1; k >= 0; k--) {
          dp[i + 1][j][k] = (dp[i + 1][j][k] +
                             (long long int)dp[i][j][k] * (f(n, j, k) - i)) %
                            MOD;
          if (j)
            dp[i + 1][j - 1][k] =
                (dp[i + 1][j - 1][k] +
                 (long long int)dp[i][j][k] * (f(n, j - 1, k) - f(n, j, k))) %
                MOD;
          if (k)
            dp[i + 1][j][k - 1] =
                (dp[i + 1][j][k - 1] +
                 (long long int)dp[i][j][k] * (f(n, j, k - 1) - f(n, j, k))) %
                MOD;
        }
      }
    }
    cout << dp[n][0][0];
  }
}
