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 mod = 1000000007;
int dp[1000005][20][2];
int fun(int n, int x, int y) {
  n /= (1LL << x);
  if (y > 0) n /= 3;
  return n;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int power = log2(n);
  dp[1][power][0] = 1;
  if ((1LL << (power - 1)) * 3 <= n) dp[1][power - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int x = 0; x <= power; x++) {
      for (int y = 0; y <= 1; y++) {
        dp[i][x][y] =
            (dp[i][x][y] + 1LL * dp[i - 1][x][y] * (fun(n, x, y) - i + 1)) %
            mod;
        if (x >= 1)
          dp[i][x - 1][y] =
              (dp[i][x - 1][y] +
               1LL * dp[i - 1][x][y] * (fun(n, x - 1, y) - fun(n, x, y))) %
              mod;
        if (y >= 1)
          dp[i][x][y - 1] =
              (dp[i][x][y - 1] +
               1LL * dp[i - 1][x][y] * (fun(n, x, y - 1) - fun(n, x, y))) %
              mod;
      }
    }
  }
  cout << dp[n][0][0] << '\n';
  return 0;
}
