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 N, f[1000001][20][2], x, y;
int cnt(int x, int y) { return N / (1 << x) / (y ? 3 : 1); }
int main() {
  cin >> N;
  while ((1 << x + 1) <= N) x++;
  f[1][x][0] = 1;
  if ((1 << x - 1) * 3 <= N) f[1][x - 1][1] = 1, y = 1;
  for (int i = 2; i <= N; i++)
    for (int j = 0; j <= x; j++)
      for (int k = 0; k <= y; k++) {
        f[i][j][k] = 1ll * f[i - 1][j][k] * (cnt(j, k) - i + 1) % 1000000007;
        if (j < x)
          f[i][j][k] = (f[i][j][k] + 1ll * f[i - 1][j + 1][k] *
                                         (cnt(j, k) - cnt(j + 1, k))) %
                       1000000007;
        if (k < y)
          f[i][j][k] = (f[i][j][k] + 1ll * f[i - 1][j][k + 1] *
                                         (cnt(j, k) - cnt(j, k + 1))) %
                       1000000007;
      }
  cout << f[N][0][0] << endl;
  return 0;
}
