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;
int n, dp[1000005][21][2];
int f(int x, int y) {
  int tmp = (1 << x);
  if (y) tmp *= 3;
  return n / tmp;
}
int main() {
  scanf("%d", &n);
  int p = 0;
  while ((1 << p) <= n) p++;
  p--;
  dp[1][p][0] = 1;
  if ((1 << (p - 1)) * 3 <= n) dp[1][p - 1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int x = 0; x <= p; x++) {
      for (int y = 0; y <= 1; y++) {
        dp[i + 1][x][y] =
            (dp[i + 1][x][y] + 1LL * dp[i][x][y] * (f(x, y) - i)) % 1000000007;
        if (x)
          dp[i + 1][x - 1][y] = (dp[i + 1][x - 1][y] +
                                 1LL * dp[i][x][y] * (f(x - 1, y) - f(x, y))) %
                                1000000007;
        if (y)
          dp[i + 1][x][y - 1] = (dp[i + 1][x][y - 1] +
                                 1LL * dp[i][x][y] * (f(x, y - 1) - f(x, y))) %
                                1000000007;
      }
    }
  }
  printf("%d", dp[n][0][0]);
}
