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>
inline int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
  return data * w;
}
const int N(1e6 + 10), LogN(21), Mod(1e9 + 7);
void Add(int &x, int y) { x = (x + y) % Mod; }
int n, m, f[N][LogN][2];
inline int cnt(int x) { return n / x; }
int main() {
  n = read(), m = log2(n), f[1][m][0] = 1;
  if (((1 << (m - 1)) * 3) <= n) f[1][m - 1][1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      Add(f[i][j][0], 1ll * f[i - 1][j][0] * (cnt(1 << j) - i + 1) % Mod);
      Add(f[i][j][0],
          1ll * f[i - 1][j + 1][0] * (cnt(1 << j) - cnt(1 << (j + 1))) % Mod);
      Add(f[i][j][0],
          1ll * f[i - 1][j][1] * (cnt(1 << j) - cnt((1 << j) * 3)) % Mod);
      Add(f[i][j][1], 1ll * f[i - 1][j][1] * (cnt(3 * (1 << j)) - i + 1) % Mod);
      Add(f[i][j][1], 1ll * f[i - 1][j + 1][1] *
                          (cnt((1 << j) * 3) - cnt((1 << (j + 1)) * 3)) % Mod);
    }
  printf("%d\n", (f[n][0][0] + Mod) % Mod);
  return 0;
}
