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;
const int mod = 1e9 + 7;
int n, dp[N][21][2];
int f[21][2];
void predeal(int n) {
  int base = 1;
  for (int i = 0; i <= 20; i++) {
    f[i][0] = n / base;
    f[i][1] = n / (base * 3);
    base <<= 1;
  }
}
int main(void) {
  scanf("%d", &n);
  predeal(n);
  int X = log2(n);
  dp[1][X][0] = 1;
  if ((1 << (X - 1)) * 3 <= n) dp[1][X - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= X; j++) {
      for (int k = 0; k <= 1; k++) {
        if (f[j][k] >= i - 1)
          dp[i][j][k] = (dp[i][j][k] +
                         (long long)dp[i - 1][j][k] * (f[j][k] - i + 1) % mod) %
                        mod;
        if (j)
          dp[i][j - 1][k] =
              (dp[i][j - 1][k] +
               (long long)dp[i - 1][j][k] * (f[j - 1][k] - f[j][k]) % mod) %
              mod;
        if (k)
          dp[i][j][k - 1] =
              (dp[i][j][k - 1] +
               (long long)dp[i - 1][j][k] * (f[j][k - 1] - f[j][k]) % mod) %
              mod;
      }
    }
  }
  printf("%d\n", dp[n][0][0]);
}
