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 MAX_N = 1e6 + 7, K = 21, M = 1e9 + 7;
long long n;
signed dp[MAX_N][K][2];
long long get(long long x) {
  if (x > n) return 0;
  return (n / x);
}
signed main() {
  memset(dp, 0, sizeof(dp));
  cin >> n;
  long long poww = 0, cur = 1;
  while (cur * 2 <= n) {
    cur *= 2;
    poww++;
  }
  dp[1][poww][0] = 1;
  if ((cur / 2) * 3 <= n) dp[1][poww - 1][1] = 1;
  for (long long i = 1; i < n; i++) {
    for (long long j = 0; j <= poww; j++) {
      for (long long k = 0; k <= 1; k++) {
        long long t = 1;
        if (k == 1) t = 3;
        dp[i + 1][j][k] =
            (dp[i + 1][j][k] + dp[i][j][k] * (get((1 << j) * t) - i)) % M;
        if (j > 0)
          dp[i + 1][j - 1][k] =
              (dp[i + 1][j - 1][k] +
               dp[i][j][k] * (get((1 << (j - 1)) * t) - get((1 << j) * t))) %
              M;
        if (k == 1)
          dp[i + 1][j][0] = (dp[i + 1][j][0] +
                             dp[i][j][k] * (get(1 << j) - get((1 << j) * t))) %
                            M;
      }
    }
  }
  cout << dp[n][0][0] << endl;
  return 0;
}
