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. 
//package codeforces;

import java.util.Scanner;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.TreeMap;
import java.util.TreeSet;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Set;

public class q {
	// Integer.bitCount(i); count no of 1 bit in a given number
	static Scanner scn = new Scanner(System.in);
	static int mod=1000000007;

	public static void main(String[] args) {
		// System.out.format("%.10f", ans);char c = sc.next().charAt(0);
		// code
		
			int n = scn.nextInt();
			long ans=1;
			for(int i=1;i<=n;i++)
			{	
				ans= (ans*i)%mod;
			}
			long uni=4;
			for(int i=4;i<=n;i++)
			{
				uni=(uni*2)%mod;
			}
			System.out.println((ans-uni+mod)%mod);
		
	}

	public static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	public static boolean[] sieveOfEratosthenes(int n) {

		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i < n; i++)
			prime[i] = true;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime
			if (prime[p] == true) {
				// Update all multiples of p
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		for (int i = 2; i <= n; i++) {
			if (prime[i] == true) {
				// list.add(i);
			}

		}
		return prime;
	}

	public static class Pair implements Comparable {
		int x;
		int y;
		int z;

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

		public int compareTo(Object o) {
			Pair pp = (Pair) o;
			if (pp.x == x)
				return 0;
			else if (x > pp.x)
				return 1;
			else
				return -1;
		}

	}

}
