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

public class CP{
  public static OutputStream out=new BufferedOutputStream(System.out);
  static Scanner sc=new Scanner(System.in);
  static long mod=1000000007l;
  static int INF=1000000;

  //nl-->neew line; //l-->line;  //arp-->array print;  //arpnl-->array print new line
  public static void nl(Object o) throws IOException{out.write((o+"\n").getBytes()); }
  public static void l(Object o) throws IOException{out.write((o+"").getBytes());}
  public static void arp(int[] o) throws IOException{for(int i=0;i<o.length;i++) out.write((o[i]+" ").getBytes()); out.write(("\n").getBytes());}
  public static void arpnl(int[] o) throws IOException{for(int i=0;i<o.length;i++) out.write((o[i]+"\n").getBytes());}
  public static void scanl(long[] a,int n) {for(int i=0;i<n;i++) a[i]=sc.nextLong();}
  public static void scani(int[] a,int n) {for(int i=0;i<n;i++) a[i]=sc.nextInt();}
  public static void scan2D(int[][] a,int n,int m) {for(int i=0;i<n;i++) for(int j=0;j<m;j++) a[i][j]=sc.nextInt();}
  //
  static long cnt;
  static TreeSet<Integer> ans;
  static int n,m;
  static boolean[][] dp,vis;
  static HashMap<Pair,Boolean> arrl;
  static Stack<Integer> stk;
  static Queue<Integer> queue;
  static HashSet<Integer> mtrnode;
  static ArrayList<Pair> arrlans;


  public static void main(String[] args) throws IOException{
    long sttm=System.currentTimeMillis();
    long mod=1000000007l;

    int n=sc.nextInt();
    long ans=1, p=1;
    for(int i=2;i<=n;i++){
        ans=ans*i%mod;
        p=2*p%mod;
    }
    nl((ans-p+mod)%mod);

    out.flush();
  }
}

class Pair{
  int x; int y;
  Pair(int x,int y){
    this.x=x;
    this.y=y;
  }
}
