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 N = 1e6 + 10;
const int mod = 1e9 + 7;
int f[N][20][2], n;
int cnt(int x) { return n / x; }
void _add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d", &n);
  int t = log2(n);
  f[1][t][0] = 1;
  f[1][t - 1][1] = (1 << (t - 1)) * 3 <= n ? 1 : 0;
  int po[2];
  po[0] = 1;
  po[1] = 3;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= t; j++) {
      for (int k = 0; k <= 1; k++) {
        if (f[i][j][k]) {
          _add(f[i + 1][j][k],
               f[i][j][k] * 1ll * (cnt((1 << j) * po[k]) - i) % mod);
          if (j)
            _add(f[i + 1][j - 1][k],
                 f[i][j][k] * 1ll *
                     (cnt((1 << (j - 1)) * po[k]) - cnt((1 << j) * po[k])) %
                     mod);
          if (k)
            _add(f[i + 1][j][k - 1],
                 f[i][j][k] * 1ll *
                     (cnt((1 << j) * po[k - 1]) - cnt((1 << j) * po[k])) % mod);
        }
      }
    }
  }
  cout << f[n][0][0];
}
