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>
namespace IO {
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
  return;
}
template <typename T>
inline void write(T x = 0) {
  T wr[51];
  wr[0] = 0;
  if (x < 0) putchar('-'), x = -x;
  if (!x) putchar(48);
  while (x) wr[++wr[0]] = x % 10, x /= 10;
  while (wr[0]) putchar(48 + wr[wr[0]--]);
  return;
}
}  // namespace IO
using IO::read;
using IO::write;
using namespace std;
const int N = 1e6 + 11, mod = 1e9 + 7;
int n, t;
int f[N][21][2];
int calc(int x, int y) {
  int ans = n >> x;
  if (y) ans /= 3;
  return ans;
}
int main() {
  read(n);
  for (t = 1; (1 << (t + 1)) <= n; ++t)
    ;
  f[1][t][0] = 1;
  if ((1 << (t - 1)) * 3 <= n) f[1][t - 1][1] = 1;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 0; j <= t; ++j) {
      for (register int k = 0; k <= 1; ++k) {
        f[i + 1][j][k] =
            (f[i + 1][j][k] + 1ll * f[i][j][k] * (calc(j, k) - i)) % mod;
        if (j)
          f[i + 1][j - 1][k] =
              (f[i + 1][j - 1][k] +
               1ll * f[i][j][k] * (calc(j - 1, k) - calc(j, k))) %
              mod;
        if (k)
          f[i + 1][j][k - 1] =
              (f[i + 1][j][k - 1] +
               1ll * f[i][j][k] * (calc(j, k - 1) - calc(j, k))) %
              mod;
      }
    }
  }
  cout << f[n][0][0] << endl;
  return 0;
}
