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.*;
public class HelloWorld{
     static long M=1_000_000_007;
     static long factorial(int n){
         long fact=1l;
         for(int i=1;i<=n;i++){
             fact=(M+(fact*i)%M)%M;
         }
         return fact;
     } 
     static long power(int n){
         long ans=1;
         long base=2;
         while(n>0){
            if((n&1)==1)
                ans=(M+(ans*base)%M)%M;
            base=(M+(base*base)%M)%M;
            n=n>>1;
             
         }
         return ans;
     }
     public static void main(String []args){
        FastReader fs=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int n=fs.nextInt();
        long fact=factorial(n),pow=power(n-1),ans=(M+(fact-pow))%M;
        //out.println(fact+" "+pow);
        out.println(ans);
        out.flush();
     }
}
class FastReader{
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st=new StringTokenizer("");
    String next(){
        while(!st.hasMoreTokens()){
            try{
                st=new StringTokenizer(br.readLine());
            }catch(IOException e){
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    int nextInt(){
        return Integer.parseInt(next());
    }
    long nextLong(){
        return Long.parseLong(next());
    }
    int[] nextIntArray(int n){
        int a[]=new int[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    }
    long[] nextLongArray(int n){
        long a[]=new long[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    }
    char[] nextCharArray(){
        return next().toCharArray();
    }
}