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 my_std {
using namespace std;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    f ^= (c == '-');
    c = getchar();
  }
  for (; c <= '9' && c >= '0'; c = getchar()) ans = ans * 10 + c - '0';
  return f ? ans : -ans;
}
template <typename _Tp>
inline int chkmin(_Tp &x, _Tp y) {
  return x > y ? x = y, 1 : 0;
}
template <typename _Tp>
inline int chkmax(_Tp &x, _Tp y) {
  return x < y ? x = y, 1 : 0;
}
const int mod = 1000000007, N = 1000010;
inline void inc(int &x, const int &y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline int ksm(int x, long long y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
template <typename _Tp>
inline _Tp gcd(_Tp x, _Tp y) {
  return y ? gcd(y, x % y) : x;
}
}  // namespace my_std
using namespace my_std;
int n, f[N][24][2], len;
int main() {
  n = read();
  while ((1 << (len + 1)) <= n) ++len;
  f[1][len][0] = 1;
  f[1][len - 1][1] = ((1 << (len - 1)) * 3) <= n;
  for (register int i = (2); i <= (n); ++i) {
    for (register int j = (0); j <= (len); ++j) {
      inc(f[i][j][0], 1ll * f[i - 1][j][0] *
                          ((n / ((1 << (j)) * ((0) ? 3 : 1))) - (i - 1)) % mod);
      inc(f[i][j][1], 1ll * f[i - 1][j][1] *
                          ((n / ((1 << (j)) * ((1) ? 3 : 1))) - (i - 1)) % mod);
      if (j < len) {
        inc(f[i][j][0], 1ll * f[i - 1][j + 1][0] *
                            ((n / ((1 << (j)) * ((0) ? 3 : 1))) -
                             (n / ((1 << (j + 1)) * ((0) ? 3 : 1)))) %
                            mod);
        inc(f[i][j][1], 1ll * f[i - 1][j + 1][1] *
                            ((n / ((1 << (j)) * ((1) ? 3 : 1))) -
                             (n / ((1 << (j + 1)) * ((1) ? 3 : 1)))) %
                            mod);
      }
      inc(f[i][j][0], 1ll * f[i - 1][j][1] *
                          ((n / ((1 << (j)) * ((0) ? 3 : 1))) -
                           (n / ((1 << (j)) * ((1) ? 3 : 1)))) %
                          mod);
    }
  }
  return printf("%d\n", f[n][0][0]), 0;
}
