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 MyClass {
    static long power(long x, long y, long p) { 
        long res = 1; // Initialize result 
        x = x % p; // Update x if it is more than or 
        // equal to p 
        while (y > 0) { 
            // If y is odd, multiply x with result 
            if (y % 2 == 1) { 
                res = (res * x) % p; 
            } 
  
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        } 
        return res; 
    } 
  
// Function to find modular 
// inverse of a under modulo p 
// using Fermat's method.  
// Assumption: p is prime 
static long modInverse(int a, long p) 
{ 
    return power(a, p - 2, p); 
}
  static int largestPower(int n, int p) { 
        // Initialize result 
        int x = 0; 
  
        // Calculate x = n/p + n/(p^2) + n/(p^3) + .... 
        while (n > 0) { 
            n /= p; 
            x += n; 
        } 
        return x; 
    } 
  
// Returns n! % p using  
// Wilson's Theorem 
static long modFact(int n, long p) { 
        if (n >= p) { 
            return 0; 
        } 
  
        long res = 1; 
  
        // Use Sieve of Eratosthenes to find all primes 
        // smaller than n 
        boolean isPrime[] = new boolean[n + 1]; 
        Arrays.fill(isPrime, true); 
        for (int i = 2; i * i <= n; i++) { 
            if (isPrime[i]) { 
                for (int j = 2 * i; j <= n; j += i) { 
                    isPrime[j] = false; 
                } 
            } 
        } 
  
        // Consider all primes found by Sieve 
        for (int i = 2; i <= n; i++) { 
            if (isPrime[i]) { 
                // Find the largest power of prime 'i' that divides n 
                int k = largestPower(n, i); 
  
                // Multiply result with (i^k) % p 
                res = (res * power(i, k, p)) % p; 
            } 
        } 
        return res; 
    } 


static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                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()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
    public static void main(String args[]) 
{
	FastReader s=new FastReader();int t=0,n=0;long ans=0;long p=1000000007;//,ans=0;
     //Scanner in=new Scanner(System.in);
     //if(in.hasNext())
     //t=s.nextInt();
     //for(int j=0;j<t;j++)
     //{
      //   if(in.hasNext())
         n=s.nextInt();
         //System.out.println("n="+modFact(n,p)+" "+n);
         ans=(modFact(n,p)-power(2,n-1,p))%p;
         
         System.out.println((ans<0)?ans+=p:ans);
     //}
    }
}