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(int f = 1, int x = 0, char ch = ' ') {
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
const int N = 1e6 + 5, P = 1e9 + 7;
int n, m, f[2][22][N];
int log(int n) { return n == 1 ? 0 : log(n >> 1) + 1; }
long long w(int i) { return n / i; }
int main() {
  n = read(), m = log(n), f[0][m][1] = 1,
  f[1][m - 1][1] = (1 << (m - 1)) * 3 <= n;
  for (int i = 2; i <= n; ++i)
    for (int j = 0; j <= m; ++j) {
      f[0][j][i] = f[0][j][i - 1] * (w(1 << j) - (i - 1) + P) % P;
      f[0][j][i] = (f[0][j][i] +
                    f[0][j + 1][i - 1] * (w(1 << j) - w(1 << (j + 1))) % P) %
                   P;
      f[0][j][i] =
          (f[0][j][i] + f[1][j][i - 1] * (w(1 << j) - w((1 << j) * 3)) % P) % P;
      f[1][j][i] = f[1][j][i - 1] * (w((1 << j) * 3) - (i - 1) + P) % P;
      f[1][j][i] =
          (f[1][j][i] +
           f[1][j + 1][i - 1] * (w((1 << j) * 3) - w((1 << (j + 1)) * 3)) % P) %
          P;
    }
  printf("%d\n", f[0][0][n]);
  return 0;
}
