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 maxn = 1e6 + 10;
const int mod = 1e9 + 7;
int n, t;
int f[maxn][20][2];
inline int power(int x, int k) {
  register int res = 1;
  while (k) {
    if (k & 1) res *= x;
    x *= x;
    k >>= 1;
  }
  return res;
}
inline int cnt(int x) { return n / x; }
int main() {
  scanf("%d", &n);
  t = log2(n);
  f[1][t][0] = 1;
  if ((1 << (t - 1)) * 3 <= n) f[1][t - 1][1] = 1;
  for (register int i = 2; i <= n; i++)
    for (register int j = 0; j <= t; j++)
      for (register int k = 0; k < 2; k++) {
        f[i][j][k] =
            (f[i][j][k] + 1ll * f[i - 1][j][k] *
                              (cnt(power(2, j) * power(3, k)) - (i - 1))) %
            mod;
        if (j < t)
          f[i][j][k] = (f[i][j][k] + 1ll * f[i - 1][j + 1][k] *
                                         (cnt(power(2, j) * power(3, k)) -
                                          cnt(power(2, j + 1) * power(3, k)))) %
                       mod;
        if (k < 1)
          f[i][j][k] = (f[i][j][k] + 1ll * f[i - 1][j][k + 1] *
                                         (cnt(power(2, j) * power(3, k)) -
                                          cnt(power(2, j) * power(3, k + 1)))) %
                       mod;
      }
  printf("%d", f[n][0][0]);
  return 0;
}
