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 MAX = 1e6 + 3, MX = 20, MOD = 1e9 + 7;
int dp[MAX][MX][3], n, pot2[MX];
void precalc() {
  pot2[0] = 1;
  for (int i = 1; i < MX; i++) pot2[i] = pot2[i - 1] * 2;
}
long long calc(int x, int y) {
  long long tmp = 1ll;
  if (y) tmp = 3;
  return n / tmp / pot2[x];
}
int main() {
  precalc();
  scanf("%d", &n);
  int m = n, log = 0;
  while (m / 2) m /= 2, log++;
  dp[1][log][0] = 1;
  if (pot2[log - 1] * 3ll <= n) dp[1][log - 1][1] = 1;
  for (int k = 2; k <= n; k++) {
    for (int i = log; i >= 0; i--) {
      for (int j = 1; j >= 0; j--) {
        dp[k][i][j] = (dp[k][i][j] +
                       1ll * dp[k - 1][i][j] * max(calc(i, j) - k + 1, 0ll)) %
                      MOD;
        if (i + 1 <= log)
          dp[k][i][j] = (dp[k][i][j] + 1ll * dp[k - 1][i + 1][j] *
                                           (calc(i, j) - calc(i + 1, j))) %
                        MOD;
        if (j + 1 < 2)
          dp[k][i][j] = (dp[k][i][j] + 1ll * dp[k - 1][i][j + 1] *
                                           (calc(i, j) - calc(i, j + 1))) %
                        MOD;
      }
    }
  }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
