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 maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
long long dp[2][22][2];
long long cal(int n, int j, int k) {
  int t = (1 << j);
  if (k) t *= 3;
  return 1ll * n / t;
}
int main() {
  int n;
  cin >> n;
  int p = 0, tmp = n;
  while (tmp) tmp >>= 1, p++;
  p--;
  int now = 1, last = 0;
  dp[1][p][0] = 1;
  if (3 * (1 << (p - 1)) <= n) dp[1][p - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    swap(now, last);
    for (int j = 0; j <= p; j++) {
      for (int k = 0; k < 2; k++) {
        dp[now][j][k] = 0;
        dp[now][j][k] =
            (dp[now][j][k] + dp[last][j][k] * (cal(n, j, k) - i + 1) % mod);
        dp[now][j][k] =
            (dp[now][j][k] +
             dp[last][j + 1][k] * (cal(n, j, k) - cal(n, j + 1, k)) % mod);
        dp[now][j][k] =
            (dp[now][j][k] +
             dp[last][j][k + 1] * (cal(n, j, k) - cal(n, j, k + 1)) % mod);
      }
    }
  }
  cout << dp[now][0][0] << "\n";
  return 0;
}
