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;
const double PI = 3.14159265358979323846;
const long long mod = 1e9 + 7;
bool check_prime(int n);
bool check_palindrone(string b);
set<int> gen_factors(int a);
set<int> gen_pfactors(int a);
set<int> gen_primes(int a);
long long combination(int a, int b, long long mod);
long long factorial(int a, long long mod);
set<int> primes = gen_primes(2);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int a;
  cin >> a;
  long long ans = factorial(a, mod);
  long long power = 1;
  a--;
  while (a--) {
    power *= 2;
    power %= mod;
  }
  ans -= power;
  ans %= mod;
  cout << ans;
}
long long combination(int a, int b, long long mod) {
  map<int, int> factors;
  if (b > a) return 0;
  for (int z : primes) {
    if (z > a) break;
    int d = a;
    while (d > 0) {
      factors[z] += d / z;
      d /= z;
    }
    d = b;
    while (d > 0) {
      factors[z] -= d / z;
      d /= z;
    }
    d = a - b;
    while (d > 0) {
      factors[z] -= d / z;
      d /= z;
    }
  }
  long long ans = 1;
  for (auto z : factors) {
    int exponent = z.second;
    while (exponent--) {
      ans *= z.first;
      if (mod != -1) ans %= mod;
    }
  }
  return ans;
}
bool check_prime(int n) {
  if (n < 2) return 0;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
bool check_palindrone(string b) {
  int length = b.length();
  for (int i = 0; i < (length + 1) / 2; i++) {
    if (b[i] != b[length - (i + 1)]) return false;
  }
  return true;
}
set<int> gen_factors(int a) {
  set<int> factors;
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      factors.insert(i);
      factors.insert(a / i);
    }
  }
  return factors;
}
set<int> gen_pfactors(int a) {
  set<int> pfactors;
  for (int z : primes) {
    if (a == 1) break;
    if (a % z == 0) {
      pfactors.insert(z);
      while (a % z == 0) a /= z;
    }
  }
  if (a != 1) pfactors.insert(a);
  return pfactors;
}
set<int> gen_primes(int a) {
  set<int> primes;
  for (int i = 2; i <= a; i++) {
    if (check_prime(i)) primes.insert(i);
  }
  return primes;
}
long long factorial(int a, long long mod) {
  long long ans = 1;
  while (a > 0) {
    ans *= a;
    if (mod > 0) ans %= mod;
    a--;
  }
  return ans;
}
