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 NMAX = 1e6;
const int LGMAX = 20;
const int MOD = 1e9 + 7;
int n;
int dp[2][LGMAX + 1][2];
int main() {
  scanf("%d", &n);
  int lg2 = 0;
  while ((1 << lg2) <= n) {
    lg2++;
  }
  lg2--;
  dp[1][lg2][0] = 1;
  if (n >= 3 && (3 << (lg2 - 1) <= n)) {
    dp[1][lg2 - 1][1] = 1;
  }
  for (int i = 1, l = 1; i < n; i++, l ^= 1) {
    memset(dp[1 ^ l], 0, sizeof(dp[1 ^ l]));
    for (int j = 0; j <= LGMAX; j++) {
      dp[1 ^ l][j][0] =
          (dp[1 ^ l][j][0] + 1LL * dp[l][j][0] * ((n / (1 << j)) - i)) % MOD;
      if (j > 0) {
        dp[1 ^ l][j - 1][0] =
            (dp[1 ^ l][j - 1][0] +
             1LL * dp[l][j][0] * ((n / (1 << (j - 1))) - (n / (1 << j)))) %
            MOD;
      }
      dp[1 ^ l][j][1] =
          (dp[1 ^ l][j][1] + 1LL * dp[l][j][1] * ((n / (3 << j)) - i)) % MOD;
      if (j > 0) {
        dp[1 ^ l][j - 1][1] =
            (dp[1 ^ l][j - 1][1] +
             1LL * dp[l][j][1] * ((n / (3 << (j - 1))) - (n / (3 << j)))) %
            MOD;
      }
      dp[1 ^ l][j][0] =
          (dp[1 ^ l][j][0] +
           1LL * dp[l][j][1] * ((n / (1 << j)) - (n / (3 << j)))) %
          MOD;
    }
  }
  printf("%d\n", dp[n & 1][0][0]);
  return 0;
}
