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;
long long int C(int n, int x, int y) {
  int den = (1 << x);
  if (y) den = den * 3;
  return n / den;
}
long long int dp[2][21][2];
int main() {
  int n;
  scanf("%d", &n);
  int x = 0;
  while ((1 << (x + 1)) <= n) x++;
  dp[n & 1][x][0] = 1;
  if ((3 << (x - 1)) <= n) dp[n & 1][x - 1][1] = 1;
  for (int p = n - 1; p >= 1; p--) {
    bool pos = p & 1;
    for (int pw2 = 0; pw2 <= x; pw2++) {
      dp[pos][pw2][0] = dp[pos ^ 1][pw2][0] * (C(n, pw2, 0) - (n - p));
      dp[pos][pw2][0] +=
          dp[pos ^ 1][pw2 + 1][0] * (C(n, pw2, 0) - C(n, pw2 + 1, 0));
      dp[pos][pw2][0] += dp[pos ^ 1][pw2][1] * (C(n, pw2, 0) - C(n, pw2, 1));
      dp[pos][pw2][0] %= 1000000007;
      dp[pos][pw2][1] = dp[pos ^ 1][pw2][1] * (C(n, pw2, 1) - (n - p));
      dp[pos][pw2][1] +=
          dp[pos ^ 1][pw2 + 1][1] * (C(n, pw2, 1) - C(n, pw2 + 1, 1));
      dp[pos][pw2][1] %= 1000000007;
    }
  }
  printf("%lld", dp[1][0][0]);
}
