A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Consider a permutation p of length n, we build a graph of size n using it as follows: 

  * For every 1 ≤ i ≤ n, find the largest j such that 1 ≤ j < i and p_j > p_i, and add an undirected edge between node i and node j 
  * For every 1 ≤ i ≤ n, find the smallest j such that i < j ≤ n and p_j > p_i, and add an undirected edge between node i and node j 



In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.

For clarity, consider as an example n = 4, and p = [3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).

A permutation p is cyclic if the graph built using p has at least one simple cycle. 

Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 10^9+7.

Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3 ≤ n ≤ 10^6).

Output

Output a single integer 0 ≤ x < 10^9+7, the number of cyclic permutations of length n modulo 10^9+7.

Examples

Input


4


Output


16

Input


583291


Output


135712853

Note

There are 16 cyclic permutations for n = 4. [4,2,1,3] is one such permutation, having a cycle of length four: 4 → 3 → 2 → 1 → 4.

Nodes v_1, v_2, …, v_k form a simple cycle if the following conditions hold: 

  * k ≥ 3. 
  * v_i ≠ v_j for any pair of indices i and j. (1 ≤ i < j ≤ k) 
  * v_i and v_{i+1} share an edge for all i (1 ≤ i < k), and v_1 and v_k share an edge. 
#include <bits/stdc++.h>
using namespace std;
const double pi = 3.14159265358979323846;
long long binpow(long long, long long);
long long mult(long long, long long);
long long add(long long, long long);
long long division(long long, long long);
long long nCr(long long, long long);
long long inv(long long);
void calc();
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  if (!v.empty()) {
    os << v.front();
    for (int i = 1; i < v.size(); ++i) os << ' ' << v[i];
  }
  return os;
}
const int N = 1000005;
long long fact[N];
void solve() {
  long long n;
  cin >> n;
  cout << (fact[n] - binpow(2, n - 1) + 1000000007) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  calc();
  for (long long i = 0; i < t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}
long long add(long long a, long long b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
  while (a < 0) a += 1000000007;
  return a;
}
long long mult(long long a, long long b) { return (a * 1ll * b) % 1000000007; }
long long binpow(long long a, long long b) {
  long long c = 1;
  while (b > 0) {
    if (b % 2 == 1) c = mult(c, a);
    a = mult(a, a);
    b /= 2;
  }
  return c;
}
long long inv(long long a) { return binpow(a, 1000000007 - 2); }
long long division(long long a, long long b) { return mult(a, inv(b)); }
void calc() {
  fact[0] = 1;
  for (long long i = 1; i <= N - 1; i++)
    fact[i] = (i * 1ll * fact[i - 1]) % 1000000007;
}
