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, K = 21, P = 1e9 + 7;
int n, m, f[N][K][2], _2[K];
void fix(int& x) { x = x < P ? x : x - P; }
int cnt(int i, int j) {
  int tmp = _2[i];
  if (j) tmp *= 3;
  return n / tmp;
}
signed main() {
  cin >> n;
  int lim = 1;
  while (lim <= n) _2[m] = lim, lim <<= 1, m++;
  m--;
  lim >>= 2;
  f[1][m][0] = 1;
  if (lim * 3 <= n) f[1][m - 1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = m; j >= 0; j--) {
      for (int k = 1; k >= 0; k--)
        if (f[i][j][k]) {
          if (j)
            fix(f[i + 1][j - 1][k] +=
                1ll * f[i][j][k] * (cnt(j - 1, k) - cnt(j, k) + P) % P);
          if (k)
            fix(f[i + 1][j][k - 1] +=
                1ll * f[i][j][k] * (cnt(j, k - 1) - cnt(j, k) + P) % P);
          fix(f[i + 1][j][k] += 1ll * f[i][j][k] * (cnt(j, k) - i + P) % P);
        }
    }
  }
  cout << f[n][0][0] << endl;
  return 0;
}
