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 long long p = 1e9 + 7;
long long ans, fact = 1, n, per[33], pre[33], m, top;
long long inv(long long x) {
  return x == 1 ? 1 : p - (p / x) * (inv(p % x)) % p;
}
long long calc(long long x) { return (x == 0) ? 0 : n / x; }
signed main() {
  cin >> n;
  m = log2(n);
  for (long long i = 1; i <= n; ++i) fact = fact * i % p;
  for (long long i = 1; i <= m; ++i) per[++top] = 2;
  if ((1 << (m - 1)) * 3 <= n) per[++top] = 3;
  do {
    pre[0] = 1;
    for (long long i = 1; i <= m; ++i) pre[i] = pre[i - 1] * per[i] % p;
    long long tmp = fact;
    for (long long i = 1; i <= m; ++i)
      tmp = tmp * (calc(pre[i]) - calc(pre[i + 1])) % p *
            inv(n - calc(pre[i + 1])) % p;
    ans = (ans + tmp) % p;
  } while (next_permutation(per + 1, per + 1 + top));
  cout << ans;
}
