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, mod = 1e9 + 7, INF = 0x3f3f3f3f;
inline long long mul(long long a, long long b) { return a * b % mod; }
inline long long add(long long a, long long b) {
  a += b;
  return a >= mod ? a - mod : a;
}
int n;
inline long long cal(int x, int y) { return n / ((1 << x) * (y ? 3 : 1)); }
int f[maxn][20][3];
int main() {
  cin >> n;
  int x = 0;
  while ((1 << (x + 1)) <= n) ++x;
  f[1][x][0] = 1;
  if ((1 << (x - 1)) * 3 <= n) f[1][x - 1][1] = 1;
  for (int i = 1; i < n; ++i)
    for (int j = 0; j <= x; ++j)
      for (int k = 0; k <= 1; ++k) {
        f[i + 1][j][k] = add(f[i + 1][j][k], mul(f[i][j][k], cal(j, k) - i));
        if (j)
          f[i + 1][j - 1][k] = add(f[i + 1][j - 1][k],
                                   mul(f[i][j][k], cal(j - 1, k) - cal(j, k)));
        if (k)
          f[i + 1][j][k - 1] = add(f[i + 1][j][k - 1],
                                   mul(f[i][j][k], cal(j, k - 1) - cal(j, k)));
      }
  cout << f[n][0][0];
  return 0;
}
