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



import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class C {

    private static final int MOD = (int) (1e9 + 7);

    static class Combinations {
        long[] factorial;
        long[] facInverse;
        long[] inverse;

        void init(int n) {
            final int MAX = n + 2;
            factorial = new long[MAX];
            facInverse = new long[MAX];
            inverse = new long[MAX];
            factorial[0] = factorial[1] = 1;
            facInverse[0] = facInverse[1] = 1;
            inverse[1] = 1;
            for (int i = 2; i < MAX; i++) {
                factorial[i] = factorial[i - 1] * i % MOD;
                final long inv = inverse[i] = MOD - inverse[MOD % i] * (MOD / i) % MOD;
                facInverse[i] = facInverse[i - 1] * inv % MOD;
            }
        }

        long ncr(int n, int r) {
            if (n < r) { return 0; }
            if (n < 0 || r < 0) { return 0; }
            return factorial[n] * (facInverse[r] * facInverse[n - r] % MOD) % MOD;
        }

        long modpow(long a, long n) {
            long res = 1;
            while (n > 0) {
                if (n % 2 == 1) {
                    res = res * a % MOD;
                }
                a = a * a % MOD;
                n /= 2;
            }
            return res;
        }
    }

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int n = Integer.parseInt(in.nextLine());
        final Combinations c = new Combinations();
        c.init((int) 1e6 + 1);
        System.out.println(c.factorial[n] - c.modpow(2, n - 1));
    }
}
