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. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

#include <bits/stdc++.h>
using namespace std;
long long signed pow_(long long signed a, long long signed e,
                      long long signed m = 1000000007) {
  long long signed r = 1;
  while (e) {
    if (e & 1) r = r * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return r;
}
int inv(long long signed a) { return int(pow_(a, 1000000007 - 2, 1000000007)); }
int add(long long signed a, long long signed b) {
  return (a + b + 1000000007) % 1000000007;
}
int mul(long long signed a, long long signed b) {
  return (1000000007 + a * b % 1000000007) % 1000000007;
}
int di(long long signed a, long long signed b) { return mul(a, inv(b)); }
long long signed ceil(long long signed a, long long signed b) {
  return a / b + (a % b != 0);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& O, pair<T1, T2> a) {
  O << "( ";
  O << a.first << ", " << a.second;
  O << " )";
  return O;
}
template <typename T>
ostream& operator<<(ostream& O, vector<T> V) {
  O << "[ ";
  for (int i = 0; i < V.size(); i++)
    O << V[i] << (i == V.size() - 1 ? "" : ", ");
  O << " ]";
  return O;
}
template <typename T>
ostream& operator<<(ostream& O, set<T> S) {
  O << "{ ";
  for (T i : S) O << i << ", ";
  O << " }";
  return O;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& O, map<T1, T2> S) {
  O << "{ ";
  for (auto i : S) O << i << ", ";
  O << " }";
  return O;
}
template <typename T = int>
class print {
 public:
  vector<T> V;
  print(vector<T> v) {
    V = v;
    cout << " <" << V << "> ";
  }
  print(T* x, int N) {
    for (int i = 0; i < N; i++) V.push_back(x[i]);
    cout << " <" << V << "> ";
  }
  friend ostream& operator<<(ostream& O, print p) { return O; }
};
int arr[1000010];
int N;
int fac[1000010];
void reset();
int ncr(int n, int r) { return di(fac[n], mul(fac[r], fac[n - r])); }
int solve() {
  cin >> N;
  int s = 0;
  for (int i = 0; i < N; i++) {
    s = add(s, ncr(N - 1, i));
  }
  cout << (add(fac[N], -s) + 1000000007) % 1000000007;
  return 0;
}
void reset() {}
signed main() {
  fac[0] = 1;
  for (int i = 1; 1 > 0 ? i < (int)1e6 : i > (int)1e6; i += 1)
    fac[i] = mul(fac[i - 1], i);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
