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;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
long long mod = 1e9 + 7;
long long BinPow(long long b, int e) {
  if (!e) return 1;
  if (e & 1) return (b * BinPow(b, e - 1)) % mod;
  long long bb = BinPow(b, e / 2);
  return (bb * bb) % mod;
}
const int maxn = 1e6 + 200;
long long fact[maxn], inv[maxn];
long long Comb(int i, int N) {
  return fact[N] * inv[i] % mod * inv[N - i] % mod;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  fact[0] = inv[0] = 1;
  fact[1] = inv[1] = 1;
  for (int i = int(2); i < int(maxn); i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv[i] = BinPow(fact[i], mod - 2);
  }
  long long res = 0;
  int N;
  cin >> N;
  for (int i = 0; i < int(N); i++) {
    res += Comb(i, N - 1);
    if (res >= mod) res -= mod;
  }
  long long tres = (fact[N] - res + 10 * mod) % mod;
  cout << tres << "\n";
  return 0;
}
