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 = 1111111;
const int md = (int)1e9 + 7;
inline void add(int& a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
}
int dp[N][22][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int x = log2(n);
  dp[1][x][0] = 1;
  if ((1 << (x - 1)) * 3 <= n) {
    dp[1][x - 1][1] = 1;
  }
  auto calc = [&](int x, int y) { return n / ((1 << x) * (y == 0 ? 1 : 3)); };
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= x; j++) {
      for (int k = 0; k < 2; k++) {
        add(dp[i + 1][j][k], dp[i][j][k] * 1LL * (calc(j, k) - i) % md);
        if (j > 0) {
          add(dp[i + 1][j - 1][k],
              dp[i][j][k] * 1LL * (calc(j - 1, k) - calc(j, k)) % md);
        }
        if (k > 0) {
          add(dp[i + 1][j][k - 1],
              dp[i][j][k] * 1LL * (calc(j, k - 1) - calc(j, k)) % md);
        }
      }
    }
  }
  cout << dp[n][0][0] << '\n';
  return 0;
}
