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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double EPS = 1e-9;
const long double PI = acos(-1);
const long long INF = 1e18;
const long long mod = 1e9 + 7;
long long factorial(long long n) {
  if (n == 0) return 1;
  long long x = 1;
  for (long long i = 1; i <= n; i++) {
    x = (x * i) % mod;
  }
  return x;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y = y >> 1;
  }
  return res % mod;
}
long long modInverse(long long n) { return power(n, mod - 2); }
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return ((factorial(n) % mod) *
          (modInverse(((factorial(r) % mod) * (factorial(n - r) % mod)) % mod) %
           mod) %
          mod);
}
long long xor_1_x(long long x) {
  if (x % 4 == 0) return x;
  if (x % 4 == 1) return 1;
  if (x % 4 == 2) return x + 1;
  return 0;
}
bool prime(long long n) {
  if (n < 2) return false;
  for (long long x = 2; x * x <= n; x++) {
    if (n % x == 0) return false;
  }
  return true;
}
vector<long long> factors(long long n) {
  vector<long long> f;
  for (long x = 2; x * x <= n; x++) {
    while (n % x == 0) {
      f.push_back(x);
      n /= x;
    }
  }
  if (n > 1) f.push_back(n);
  return f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test_cases = 1;
  while (test_cases--) {
    long long n;
    cin >> n;
    long long ans = ((factorial(n) % mod) - (power(2, n - 1) % mod)) % mod;
    if (ans < 0) ans += mod;
    cout << ans << endl;
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}
