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 mod = 1e9 + 7;
const int N = 1e6;
const int M = 20;
long long dp[M][3];
long long cou[M][3];
int main() {
  int n;
  cin >> n;
  int mj = 0;
  for (int i = 0; (1 << i) <= n; i++) {
    cou[i][0] = n / (1 << i);
    cou[i][1] = n / (3 * (1 << i));
    mj = i;
  }
  dp[mj][0] = cou[mj][0];
  if (mj > 0) dp[mj - 1][1] = cou[mj - 1][1];
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= mj; j++) {
      for (int t = 0; t <= 1; ++t) {
        long long nxt = dp[j][t] * max((long long)0, (cou[j][t] - i)) % mod;
        nxt += (dp[j + 1][t] *
                max((long long)0,
                    (cou[j][t] - i - max((long long)0, cou[j + 1][t] - i))) %
                mod);
        nxt += (dp[j][t + 1] *
                max((long long)0,
                    (cou[j][t] - i - max((long long)0, cou[j][t + 1] - i))) %
                mod);
        dp[j][t] = nxt % mod;
      }
    }
  }
  cout << dp[0][0];
}
