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;
int po[21], n, po1[3];
int dp[1000005][20][2];
int calc(int x, int y) {
  long long o = po[x] * po1[y];
  return n / o;
}
int main() {
  po[0] = 1;
  po1[0] = 1;
  po1[1] = 3;
  int X;
  cin >> n;
  for (int i = 1; i <= 20; i++) {
    po[i] = po[i - 1] * 2;
    if (po[i] > n) {
      X = i - 1;
      break;
    }
  }
  dp[1][X][0] = 1;
  if (po[X - 1] * 3 <= n) dp[1][X - 1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int x = 0; x <= X; x++) {
      for (int y = 0; y <= 1; y++) {
        dp[i + 1][x][y] =
            (dp[i + 1][x][y] + 1LL * dp[i][x][y] * (calc(x, y) - i)) %
            1000000007;
        if (x > 0)
          dp[i + 1][x - 1][y] =
              (dp[i + 1][x - 1][y] +
               1LL * dp[i][x][y] * (calc(x - 1, y) - calc(x, y))) %
              1000000007;
        if (y > 0)
          dp[i + 1][x][y - 1] =
              (dp[i + 1][x][y - 1] +
               1LL * dp[i][x][y] * (calc(x, y - 1) - calc(x, y))) %
              1000000007;
      }
    }
  }
  cout << dp[n][0][0];
  return 0;
}
