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;
const int M = 1e6 + 10;
int n, dp[M][21][2];
int f(int x, int y) { return n / ((y ? 3 : 1) << x); }
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  int p = log2(n);
  dp[1][p][0] = 1;
  if ((1 << (p - 1)) * 3 <= n) dp[1][p - 1][1] = 1;
  for (int i = 1; i <= n - 1; i++)
    for (int x = 0; x <= 20; 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) % mod) % mod;
        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)) % mod) %
              mod;
        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)) % mod) %
              mod;
      }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
