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 MOD = 1e9 + 7;
int dp[1000006][20][2];
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int mul(int a, int b) {
  long long c = ((long long)a * b);
  if (c >= MOD) c %= MOD;
  return (int)c;
}
int f(int n, int j, int k) {
  int num = 1 << j;
  if (k) num *= 3;
  return n / num;
}
int main() {
  int n;
  scanf("%d", &n);
  int two = 0;
  while (1 << two <= n) two++;
  two--;
  int best = 1 << two;
  dp[1][two][0] = 1;
  if (3 * best <= n * 2) dp[1][two - 1][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= two; j++) {
      for (int k = 0; k <= 1; k++) {
        if (!dp[i][j][k]) continue;
        dp[i + 1][j][k] =
            add(dp[i + 1][j][k], mul(dp[i][j][k], max(0, f(n, j, k) - i)));
        if (j)
          dp[i + 1][j - 1][k] =
              add(dp[i + 1][j - 1][k],
                  mul(dp[i][j][k], max(0, f(n, j - 1, k) - f(n, j, k))));
        if (k)
          dp[i + 1][j][k - 1] =
              add(dp[i + 1][j][k - 1],
                  mul(dp[i][j][k], max(0, f(n, j, k - 1) - f(n, j, k))));
      }
    }
  }
  printf("%d\n", dp[n][0][0]);
  return 0;
}
