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 INF = 0x3ffffff;
const long long LL_INF = 0x3fffffffffffffffll;
const int MOD = 1e9 + 7;
const long long HASH_KEY = 6151;
const long long HASH_MOD = 1610612741;
const int MAXN = 1e6 + 3;
int dp[MAXN][21][2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int p = 0;
  while ((1 << p) <= n) {
    ++p;
  }
  auto gao = [&](const int &x, const int &y) {
    long long t = n >> x;
    return y ? t / 3 : t;
  };
  auto add = [&](int &x, const int &y) {
    x += y;
    if (x >= MOD) {
      x -= MOD;
    }
  };
  dp[1][p - 1][0] = 1;
  dp[1][p - 2][1] = gao(p - 2, 1);
  for (int i = int(1); i < int(n); ++i) {
    for (int x = int(0); x < int(p); ++x) {
      for (int y = int(0); y < int(2); ++y) {
        add(dp[i + 1][x][y], dp[i][x][y] * (gao(x, y) - i) % MOD);
        if (x) {
          add(dp[i + 1][x - 1][y],
              dp[i][x][y] * (gao(x - 1, y) - gao(x, y)) % MOD);
        }
        if (y) {
          add(dp[i + 1][x][y - 1],
              dp[i][x][y] * (gao(x, y - 1) - gao(x, y)) % MOD);
        }
      }
    }
  }
  cout << dp[n][0][0] << "\n";
  return 0;
}
