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 MAX = 1e6 + 1;
const long long inf = 1e18;
long long mod = 1e9 + 7;
int dp[MAX][20][2];
int n;
int con(int x, int y) {
  int a = (1 << x);
  if (y == 0) return n / a;
  return n / (a * 3);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int l = (int)log2(n);
  dp[1][l][0] = 1;
  if (con(l - 1, 1) > 0) dp[1][l - 1][1] = 1;
  for (int i = 2; i < n + 1; i++) {
    for (int j = 0; j < l + 1; j++) {
      for (int k = 0; k < 2; k++) {
        dp[i][j][k] = (1LL * dp[i][j][k] +
                       1LL * dp[i - 1][j][k] * (con(j, k) - (i - 1))) %
                      mod;
        if (j > 0)
          dp[i][j - 1][k] =
              (1LL * dp[i][j - 1][k] +
               1LL * dp[i - 1][j][k] * (con(j - 1, k) - con(j, k))) %
              mod;
        if (k > 0)
          dp[i][j][k - 1] =
              (1LL * dp[i][j][k - 1] +
               1LL * dp[i - 1][j][k] * (con(j, k - 1) - con(j, k))) %
              mod;
      }
    }
  }
  cout << dp[n][0][0];
  return 0;
}
