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;
long long n;
int dp[1000010][21][2];
long long cnt(long long x, long long y) {
  return n / ((1 << x) * (y == 1 ? 3 : 1));
}
int main() {
  scanf("%lld", &n);
  int bit = log2(n);
  dp[1][bit][0] = 1;
  if ((1 << bit - 1) * 3 <= n) {
    dp[1][bit - 1][1] = 1;
  }
  for (register int i = 2; i <= n; i++) {
    for (register int j = 0; j <= bit; j++) {
      for (register int k = 0; k <= 1; k++) {
        dp[i][j][k] =
            (dp[i][j][k] + 1ll * dp[i - 1][j][k] * (cnt(j, k) - i + 1) % mod) %
            mod;
        if (j < bit) {
          dp[i][j][k] = (dp[i][j][k] + 1ll * dp[i - 1][j + 1][k] *
                                           (cnt(j, k) - cnt(j + 1, k)) % mod) %
                        mod;
        }
        if (k < 1) {
          dp[i][j][k] = (dp[i][j][k] + 1ll * dp[i - 1][j][k + 1] *
                                           (cnt(j, k) - cnt(j, k + 1)) % mod) %
                        mod;
        }
      }
    }
  }
  printf("%d", dp[n][0][0]);
  return 0;
}
