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 = 1e9 + 7;
int dp[2][21][2];
int n, x;
int calc(int x, int y) {
  int tmp = 1 << x;
  if (y == 1) tmp *= 3;
  return n / tmp;
}
int main() {
  cin >> n;
  x = 0;
  while ((1 << x) <= n) x++;
  x--;
  dp[1][x][0] = 1;
  if ((1 << (x - 1)) * 3 <= n) {
    dp[1][x - 1][1] = 1;
  }
  for (int i = 1; i < n; i++) {
    for (int p2 = 0; p2 <= x; p2++) {
      for (int p3 = 0; p3 <= 1; p3++) {
        dp[(i + 1) % 2][p2][p3] =
            (dp[(i + 1) % 2][p2][p3] +
             1ll * dp[i % 2][p2][p3] * (calc(p2, p3) - i)) %
            mod;
        if (p2 > 0) {
          dp[(i + 1) % 2][p2 - 1][p3] =
              (dp[(i + 1) % 2][p2 - 1][p3] +
               1ll * dp[i % 2][p2][p3] * (calc(p2 - 1, p3) - calc(p2, p3))) %
              mod;
        }
        if (p3 > 0) {
          dp[(i + 1) % 2][p2][p3 - 1] =
              (dp[(i + 1) % 2][p2][p3 - 1] +
               1ll * dp[i % 2][p2][p3] * (calc(p2, p3 - 1) - calc(p2, p3))) %
              mod;
        }
      }
    }
    for (int p2 = 0; p2 <= x; p2++) {
      for (int p3 = 0; p3 <= 1; p3++) {
        dp[i % 2][p2][p3] = 0;
      }
    }
  }
  cout << dp[n % 2][0][0] << endl;
  return 0;
}
