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 N = 1000010, mod = 1e9 + 7;
long long n, lt;
int f[N][21][2];
inline long long read() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline long long get(long long x) { return n / x; }
int main() {
  n = read();
  lt = log2(n);
  f[1][lt][0] = 1;
  if ((1 << (lt - 1)) * 3 <= n) f[1][lt - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= lt; j++) {
      f[i][j][0] =
          (1ll * f[i - 1][j][0] * (get(1 << j) - (i - 1)) +
           1ll * f[i - 1][j + 1][0] * (get(1 << j) - get(1 << (j + 1))) +
           1ll * f[i - 1][j][1] * (get(1 << j) - get((1 << j) * 3))) %
          mod;
      f[i][j][1] = (1ll * f[i - 1][j][1] * (get((1 << j) * 3) - (i - 1)) +
                    1ll * f[i - 1][j + 1][1] *
                        (get((1 << j) * 3) - get((1 << (j + 1)) * 3))) %
                   mod;
    }
  }
  printf("%d", f[n][0][0]);
}
