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. 
    import org.omg.CORBA.MARSHAL;

    import java.util.*;
    import java.io.*;

    public class Main {

        public static void main(String[] args) throws IOException {
            Scanner sc = new Scanner(System.in);
            PrintWriter pw = new PrintWriter(System.out);
            int n = sc.nextInt();
            long[]fac= new long[n+1];
            fac[0]=fac[1]=1;
            int mod =(int)1e9+7;
            long pow=1;
            for (int i =0;i<n-1;i++){
                pow*=2;
                pow%=mod;
            }
            for (int i =1;i<=n;i++){
                fac[i]=(fac[i-1]*i)%mod;
            }
            pw.println((fac[n]-pow+mod)%mod);
            pw.close();

        }
        static class Scanner {
            BufferedReader br;
            StringTokenizer st;

            public Scanner(InputStream s) {
                br = new BufferedReader(new InputStreamReader(s));
            }

            public String next() throws IOException {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            }

            public int nextInt() throws IOException {
                return Integer.parseInt(next());
            }

            public long nextLong() throws IOException {
                return Long.parseLong(next());
            }

            public double nextDouble() throws IOException {
                return Double.parseDouble(next());
            }

            public int[] nextIntArr(int n) throws IOException {
                int[] arr = new int[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = Integer.parseInt(next());
                }
                return arr;
            }

        }

    }