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>
int main() {
  static int dp[19 + 1], dq[19 + 1], dp0[19], dq0[19], dp1[19], dq1[19];
  int n, i, j, kj, kj_, l, ans;
  scanf("%d", &n);
  l = 0;
  while (1 << (l + 1) <= n) l++;
  ans = 0;
  dp[l] = 1;
  for (i = 1; i < n; i++) {
    for (j = 0; j <= l; j++)
      if (dp[j]) {
        kj = n >> j;
        dq[j] = (dq[j] + (long long)dp[j] * ((n >> j) - i)) % 1000000007;
        if (j > 0) {
          kj_ = n >> (j - 1);
          dq[j - 1] = (dq[j - 1] + (long long)dp[j] * (kj_ - kj)) % 1000000007;
        }
      }
    for (j = 0; j <= l; j++) dp[j] = dq[j], dq[j] = 0;
  }
  ans = (ans + dp[0]) % 1000000007;
  if ((1 << (l - 1)) * 3 <= n) {
    dp1[l - 1] = 1;
    for (i = 1; i < n; i++) {
      for (j = 0; j < l; j++) {
        if (dp0[j]) {
          kj = n >> j;
          dq0[j] = (dq0[j] + (long long)dp0[j] * (kj - i)) % 1000000007;
          if (j > 0) {
            kj_ = n >> (j - 1);
            dq0[j - 1] =
                (dq0[j - 1] + (long long)dp0[j] * (kj_ - kj)) % 1000000007;
          }
        }
        if (dp1[j]) {
          kj = (n >> j) / 3;
          dq1[j] = (dq1[j] + (long long)dp1[j] * (kj - i)) % 1000000007;
          if (j > 0) {
            kj_ = (n >> (j - 1)) / 3;
            dq1[j - 1] =
                (dq1[j - 1] + (long long)dp1[j] * (kj_ - kj)) % 1000000007;
          }
          kj_ = n >> j;
          dq0[j] = (dq0[j] + (long long)dp1[j] * (kj_ - kj)) % 1000000007;
        }
      }
      for (j = 0; j < l; j++) {
        dp0[j] = dq0[j], dq0[j] = 0;
        dp1[j] = dq1[j], dq1[j] = 0;
      }
    }
    ans = (ans + dp0[0]) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}
