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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long llinf = (1LL << 62);
const int inf = (1 << 30);
const int nmax = 1e6 + 50;
const long long mod = 1e9 + 7;
using namespace std;
int n, x, i, j, t;
long long f[2][22][2];
int nr(int x, int y) {
  int rs = (1 << x);
  if (y) rs *= 3;
  return n / rs;
}
void ad(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n;
  while ((1 << x) <= n) x++;
  x--;
  f[1][x][0] = 1;
  if ((1 << (x - 1)) * 3 <= n) f[1][x - 1][1] = 1;
  for (i = 1; i < n; i++) {
    for (j = 0; j <= x; j++) {
      for (t = 0; t <= 1; t++) {
        ad(f[0][j][t], f[1][j][t] * 1LL * (nr(j, t) - i) % mod);
        if (j)
          ad(f[0][j - 1][t],
             f[1][j][t] * 1LL * (nr(j - 1, t) - nr(j, t)) % mod);
        if (t)
          ad(f[0][j][t - 1],
             f[1][j][t] * 1LL * (nr(j, t - 1) - nr(j, t)) % mod);
      }
    }
    for (j = 0; j <= x; j++) {
      for (t = 0; t <= 1; t++) {
        f[1][j][t] = f[0][j][t];
        f[0][j][t] = 0;
      }
    }
  }
  cout << f[1][0][0] << endl;
  return 0;
}
