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;
inline int read() {
  char ch = getchar();
  int res = 0, flag = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * flag;
}
int n, logn, f[1000001][21][2];
const int Mod = 1e9 + 7;
int main() {
  n = read();
  logn = log2(n);
  f[1][logn][0] = 1;
  f[1][logn - 1][1] = ((1 << (logn - 1)) * 3 <= n);
  for (register int i = 2; i <= n; ++i)
    for (register int j = 0; j <= logn; ++j) {
      f[i][j][0] =
          ((((long long)f[i - 1][j][0] * (n / (1 << j) - i + 1)) % Mod +
            ((long long)f[i - 1][j + 1][0] *
             (n / (1 << j) - n / (1 << (j + 1)))) %
                Mod) %
               Mod +
           ((long long)f[i - 1][j][1] * (n / (1 << j) - n / ((1 << j) * 3))) %
               Mod) %
          Mod;
      f[i][j][1] =
          (((long long)f[i - 1][j][1] * (n / ((1 << j) * 3) - i + 1)) % Mod +
           ((long long)f[i - 1][j + 1][1] *
            (n / ((1 << j) * 3) - n / ((1 << (j + 1)) * 3))) %
               Mod) %
          Mod;
    }
  printf("%d\n", f[n][0][0]);
  return 0;
}
