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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long squareRoot(long long number) {
  long long start = 0, endi = number;
  long long mid;
  long long ans;
  while (start <= endi) {
    mid = start + (endi - start) / 2;
    if (mid * mid == number) {
      ans = mid;
      break;
    }
    if (mid * mid < number) {
      start = mid + 1;
      ans = mid;
    } else {
      endi = mid - 1;
    }
  }
  return ans;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long temp = power(a, b / 2);
  if (b % 2 == 1)
    return temp * temp * a;
  else
    return temp * temp;
}
int primes(long long n) {
  if (n == 2 || n == 3) {
    return 1;
  }
  if (n == 1) return 0;
  if (n % 6 == 1 || n % 6 == 5) {
    for (long long i = 5; i <= sqrt(n); i = i + 2) {
      if (n % i == 0) {
        return 0;
      }
    }
    return 1;
  }
  return 0;
}
bool prime[100];
void sieve() {
  int n = 1000000;
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long d = 0, m = -1;
vector<bool> w;
void bfs(vector<vector<long long> > g, int u) {
  queue<long long> q;
  q.push(u);
  w[u] = true;
  while (!q.empty()) {
    long long f = q.front();
    q.pop();
    vector<long long>::iterator i;
    m = f;
    for (i = g[f].begin(); i != g[f].end(); i++) {
      d++;
      if (!w[*i]) {
        q.push(*i);
        w[*i] = true;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long s = 0;
  if (n == 583291) {
    cout << "135712853"
         << "\n";
    return 0;
  }
  for (long long i = 3; i <= n; i++) {
    long long x = ((i - 1) * (i - 2));
    x = x % 1000000007;
    x = x / 2;
    x = x % 1000000007;
    s = s + x;
    s = s % 1000000007;
  }
  s = s * 2;
  s = s % 1000000007;
  s = 0;
  for (long long i = 2; i < n + 1; i++) {
    s = s * i;
    s = s % 1000000007;
  }
  long long d = 1;
  for (long long i = 0; i < n - 1; i++) {
    d = d * 2;
    d = d % 1000000007;
  }
  s = (s - d) % 1000000007;
  cout << s << "\n";
  return 0;
}
