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.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main{
       public static void main(String args[]){
              FastScanner in=new FastScanner();
              long fac[]=new long[(int)1e6+1];
              fac[0]=1;
              for(int i=1;i<=1e6;i++)
                  fac[i]=(i*fac[i-1])%mod;
              
                  int n=in.nextInt();
                  long ans=fac[n]-power(2,n-1);
                  ans=(ans+mod)%mod;
                  System.out.println(ans);
              }
       public static int rec(int arr[][],int i,int j,int n){
           int sum=0;
           sum+=arr[i][j];
           
           int max=0;
          for(int k=j+1;k<n;k++)
              max=Math.max(max,rec(arr,1-i,k,n));
          
          sum+=max;
          
          return sum;
       }
      
       public static void rec(int i,int n,int count){
           if(i==n){
               count++;
               return;
           }
           if(i>n)
               return;
           
           rec(i+1,n,count);
           rec(i+2,n,count);
       }
       public static long lcm(long x, long y) {
        return ((x * y) / gcd(x, y));
    }
       public static long gcd(long a,long b){
           if(a==0)
               return b;
           
           return gcd(b%a,a);
       }
       public static int bakloli(char ch[],char c,int k){
           int n=ch.length;
           int k1=0;
           int i=0;
           int j=0;
           int ans=0;
           while(i<n){
               if(j>=n)
                   break;
               while(j<n && (ch[j]==c || k1<k)){
                   if(ch[j]!=c){
                       k1++;
                       
                   }
                   j++;
               }
               ans=Math.max(j-i,ans);
               if(ch[i]!=c)
                   k1--;
               
               i++;
           }
           return ans;
       }
       static int mod=(int)1e9+7;
       public static long power(long a,long b){
          long ans=1;
          while(b > 0) {
    		if((b&1) == 1) ans = (ans*a) % mod;
    		a = (a*a) % mod;
    		b >>= 1;
    	}
          return ans;
       }
       public static int largestfactor(int n){
          
           for(int i=2;i*i<=n;i++){
               if(n%i==0)
                   return n/i;
           }
           return 1;
       }
      
       public static int bs(int ele,int n,int arr[]){
           int low=0;
           int high=n-1;
           while(low<=high){
               int mid=(low+high)/2;
               if(arr[mid]>=ele){
                   low=mid+1;
               }
               else
                   high=mid-1;
           }
           return low;
       }
       private static int mergeAndCount(long[] arr, int l,
                                     int m, int r)
    {
 
        // Left subarray
        long[] left = Arrays.copyOfRange(arr, l, m + 1);
 
        // Right subarray
        long[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
 
        int i = 0, j = 0, k = l, swaps = 0;
 
        while (i < left.length && j < right.length) {
            if (left[i] <= right[j])
                arr[k++] = left[i++];
            else {
                arr[k++] = right[j++];
                swaps += (m + 1) - (l + i);
            }
        }
        while (i < left.length)
            arr[k++] = left[i++];
        while (j < right.length)
            arr[k++] = right[j++];
        return swaps;
    }
 
    // Merge sort function
    private static int mergeSortAndCount(long[] arr, int l,
                                         int r)
    {
 
        // Keeps track of the inversion count at a
        // particular node of the recursion tree
        int count = 0;
 
        if (l < r) {
            int m = (l + r) / 2;
 
            // Total inversion count = left subarray count
            // + right subarray count + merge count
 
            // Left subarray count
            count += mergeSortAndCount(arr, l, m);
 
            // Right subarray count
            count += mergeSortAndCount(arr, m + 1, r);
 
            // Merge count
            count += mergeAndCount(arr, l, m, r);
        }
 
        return count;
    }
}
class FastScanner {
		java.io.BufferedReader br = new java.io.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());
		}
 
		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
	}
class Pair{
    long x;
    long y;
    Pair(long x,long y){
        this.x=x;
        this.y=y;
    }
}
class Sorting implements Comparator<Pair>{
    public int compare(Pair p1,Pair p2){
       
        return (int)(p2.x-p1.x);
    }
}