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 long long N = 1e6 + 10, lg = 20, mod = 1e9 + 7;
long long n, f[lg][2], dp[2][lg][2];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < lg; i++) {
    f[i][0] = n / (1LL << i);
    f[i][1] = n / (3 * (1LL << i));
  }
  long long x = 1, cnt = 0;
  while (2 * x <= n) x *= 2, cnt++;
  dp[0][cnt][0] = 1;
  if (x / 2 * 3 <= n) dp[0][cnt - 1][1] = 1;
  for (long long i = 1; i < n; i++) {
    long long now = (i & 1), prv = 1 - now;
    for (long long j = 0; j <= cnt; j++)
      for (long long k = 0; k < 2; k++)
        dp[now][j][k] = (dp[prv][j][k] * (f[j][k] - i)) % mod;
    for (long long j = 0; j < cnt; j++)
      for (long long k = 0; k < 2; k++)
        dp[now][j][k] = (dp[now][j][k] +
                         dp[prv][j + 1][k] * (f[j][k] - f[j + 1][k]) % mod) %
                        mod;
    for (long long j = 0; j <= cnt; j++)
      dp[now][j][0] =
          (dp[now][j][0] + dp[prv][j][1] * (f[j][0] - f[j][1]) % mod) % mod;
  }
  cout << (dp[(n - 1) & 1][0][0] + mod) % mod;
}
