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;
long long dp[2][25][2];
int main() {
  int n;
  cin >> n;
  int tmp = n, sum = 0;
  while (tmp > 1) tmp /= 2, sum++;
  dp[1][sum][0] = 1;
  if ((1 << (sum - 1)) * 3 <= n) dp[1][sum - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= sum; j++) {
      dp[i & 1][j][0] = dp[i & 1][j][1] = 0;
      dp[i & 1][j][0] += dp[!(i & 1)][j][0] * (n / (1 << j) - (i - 1));
      dp[i & 1][j][1] += dp[!(i & 1)][j][1] * (n / ((1 << j) * 3) - (i - 1));
      dp[i & 1][j][0] +=
          dp[!(i & 1)][j + 1][0] * (n / (1 << j) - n / (1 << (j + 1)));
      dp[i & 1][j][1] += dp[!(i & 1)][j + 1][1] *
                         (n / ((1 << j) * 3) - n / ((1 << (j + 1)) * 3));
      dp[i & 1][j][0] +=
          dp[!(i & 1)][j][1] * (n / (1 << j) - n / ((1 << j) * 3));
      dp[i & 1][j][0] %= 1000000007;
      dp[i & 1][j][1] %= 1000000007;
    }
  }
  cout << dp[n & 1][0][0];
  return 0;
}
