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 f[1000005][20][2], cnt[20][2];
const int mod = 1e9 + 7;
int main() {
  int n, s, t;
  cin >> n;
  for (s = 1, t = 0; s <= n; s <<= 1, t++)
    ;
  s >>= 1;
  t--;
  f[1][t][0] = 1;
  if (s / 2 * 3 <= n) f[1][t - 1][1] = 1;
  for (int x = 0; x <= t; x++)
    for (int y = 0; y <= 1; y++) cnt[x][y] = n / (pow(2, x) * pow(3, y));
  for (int i = 2; i <= n; i++) {
    for (int x = 0; x <= t; x++)
      for (int y = 0; y <= 1; y++) {
        f[i][x][y] += (long long)f[i - 1][x][y] * (cnt[x][y] - i + 1) % mod;
        f[i][x][y] %= mod;
        if (x + 1 <= t)
          f[i][x][y] +=
              (long long)f[i - 1][x + 1][y] * (cnt[x][y] - cnt[x + 1][y]) % mod;
        f[i][x][y] %= mod;
        if (y + 1 <= 1)
          f[i][x][y] +=
              (long long)f[i - 1][x][y + 1] * (cnt[x][y] - cnt[x][y + 1]) % mod;
        f[i][x][y] %= mod;
      }
  }
  cout << f[n][0][0] << endl;
  return 0;
}
