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;
int n;
int f[N][20][2], cnt[22][22], p = 1e9 + 7;
void fix(int &x) {
  x %= p;
  if (x < 0) x += p;
}
int quickpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    b >>= 1;
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i <= 20; i++)
    for (int j = 0; j <= 20; j++)
      cnt[i][j] = n / (quickpow(2, i) * quickpow(3, j));
  int x = n;
  int mi = 0;
  while (x > 1) {
    x /= 2;
    mi++;
  }
  f[1][mi][0] = 1;
  if (quickpow(2, mi) / 2 * 3 <= n) f[1][mi - 1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= 19; j++) {
      for (int k = 0; k <= 1; k++) {
        if (j > 0)
          f[i + 1][j - 1][k] +=
              ((long long)f[i][j][k] * (cnt[j - 1][k] - cnt[j][k])) % p;
        if (k > 0)
          f[i + 1][j][k - 1] +=
              ((long long)f[i][j][k] * (cnt[j][k - 1] - cnt[j][k])) % p;
        if (cnt[j][k] > i)
          f[i + 1][j][k] += ((long long)f[i][j][k] * (cnt[j][k] - i)) % p;
        fix(f[i + 1][j][k - 1]);
        fix(f[i + 1][j - 1][k]);
        fix(f[i + 1][j][k]);
      }
    }
  }
  cout << f[n][0][0];
  return 0;
}
