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 = 1000005, P = 1000000007;
int n, k, dp[N][20][2], cnt[20][2];
int main() {
  scanf("%d", &n);
  while (1 << k + 1 <= n) ++k;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; !(i & ((1 << j) - 1)); ++j)
      for (int k = 0; k <= (i % 3 == 0); ++k) ++cnt[j][k];
  dp[1][k][0] = 1;
  if (k && 3 << k - 1 <= n) dp[1][k - 1][1] = 1;
  for (int i = 1; i < n; ++i)
    for (int a = 0; a <= k; ++a)
      for (int b = 0; b < 2; ++b)
        if (dp[i][a][b]) {
          if (i != cnt[a][b])
            dp[i + 1][a][b] =
                (dp[i + 1][a][b] + 1LL * dp[i][a][b] * (cnt[a][b] - i)) % P;
          if (a)
            dp[i + 1][a - 1][b] =
                (dp[i + 1][a - 1][b] +
                 1LL * dp[i][a][b] * (cnt[a - 1][b] - cnt[a][b])) %
                P;
          if (b)
            dp[i + 1][a][b - 1] =
                (dp[i + 1][a][b - 1] +
                 1LL * dp[i][a][b] * (cnt[a][b - 1] - cnt[a][b])) %
                P;
        }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
