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 <typename T>
void print(array<T, 2> &arr);
template <typename T>
void print(vector<T> &v);
template <typename T>
void print(deque<T> &v);
template <typename T>
T _gcd(T a, T b);
template <typename T>
vector<T> sieve(T n);
template <typename T>
T power(T x, T y, long long p = LLONG_MAX);
long long nxt();
template <typename T>
T modInverse(T n, T p);
long long gcd(vector<long long> &diff);
bool isPowerof2(long long x);
vector<long long> divisors(long long n);
long long nCrModPFermat(long long n, long long r, long long p = LLONG_MAX);
long long sum(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  while (T--) {
    long long n;
    cin >> n;
    long long mod = 1e9 + 7;
    long long fac = 1;
    for (int i = 1; i <= n; i++) {
      fac *= i;
      fac %= mod;
    }
    long long ans = fac - power(2ll, n - 1, mod);
    if (ans < 0) ans += mod;
    cout << ans << endl;
  }
}
template <typename T>
T power(T x, T y, long long p) {
  T res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
template <typename T>
vector<T> sieve(T n) {
  vector<bool> prime(n + 1, true);
  for (int i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (int j = 2; j * i <= n; j++) {
        prime[i * j] = false;
      }
    }
  }
  vector<T> ans;
  for (int i = 2; i <= n; i++) {
    if (prime[i]) ans.push_back(i);
  }
  return ans;
}
template <typename T>
T _gcd(T a, T b) {
  T temp1 = max(a, b), temp2 = min(a, b);
  a = temp1, b = temp2;
  if (a == 0) return b;
  if (b == 0) return a;
  if (a % b == 0) return b;
  return _gcd(b, a % b);
}
template <typename T>
void print(array<T, 2> &arr) {
  cout << arr[0] << " " << arr[1] << endl;
}
template <typename T>
void print(vector<T> &v) {
  for (T x : v) cout << x << " ";
  cout << endl;
}
template <typename T>
T modInverse(T n, T p) {
  return power(n, p - 2, p) % p;
}
template <typename T>
void print(deque<T> &v) {
  for (auto x : v) cout << x << " ";
  cout << endl;
}
long long nxt() {
  long long x;
  cin >> x;
  return x;
}
bool isPowerof2(long long x) { return !(x && (x & (x - 1))); }
vector<long long> divisors(long long n) {
  vector<long long> ans;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      ans.push_back(i);
      if (n / i != i) ans.push_back(n / i);
    }
  }
  ans.push_back(1);
  return ans;
}
long long nCrModPFermat(long long n, long long r, long long p) {
  static bool calc = 0;
  static const long long facn = 3e5;
  static const long long mod = LLONG_MAX;
  static long long fac[facn + 1];
  if (!calc) {
    fac[0] = 1;
    for (int i = 1; i <= facn; i++) fac[i] = (fac[i - 1] * i) % mod;
  }
  calc = 1;
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
