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 maxn = 1e6 + 5;
const long long mod = 1e9 + 7;
int n;
int dp[maxn][21][3];
int f(int x) {
  int res = 0;
  while (x) {
    res++;
    x >>= 1;
  }
  --res;
  return res;
}
int N(int x, int y) {
  int res = (1ll << x);
  res = res * (y ? 3 : 1);
  return res % mod;
}
int main() {
  scanf("%d", &n);
  if (n <= 2) {
    puts("1");
    return 0;
  }
  dp[1][f(n)][0] = 1;
  if (f(n / 3) + 1 >= f(n)) {
    dp[1][f(n / 3)][1] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 0; (1 << j) <= n; j++) {
      for (int k = 0; k < 2; k++) {
        if (n / N(j, k) >= i - 1) {
          dp[i][j][k] += dp[i - 1][j][k] * 1ll * (n / N(j, k) - i + 1) % mod;
        }
        dp[i][j][k] %= mod;
        dp[i][j][k] +=
            dp[i - 1][j + 1][k] * 1ll * (n / N(j, k) - n / N(j + 1, k)) % mod;
        dp[i][j][k] %= mod;
        dp[i][j][k] +=
            dp[i - 1][j][k + 1] * 1ll * (n / N(j, k) - n / N(j, k + 1)) % mod;
        dp[i][j][k] %= mod;
      }
    }
  }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
