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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


/**
 * @author b
 *
 */
public class C {
	
	
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner();
			solve(scanner);
	}
	
	static double mod = Math.pow(10, 9) + 7;
	
	private static void solve(Scanner scanner) {
		
		
		int n = scanner.nextInt();
		long n1 = n;
		long res = 1;
		long ill = 1;
		while (n1 > 1) {
			res = (long) ((res * n1) % mod);
			ill = (long) ((ill * 2) % mod);
			n1--;
		}
		
		System.out.println((long) ((res - ill + mod) % mod));
	}

	// Props to SecondThread
	static class Scanner {
		BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tokenizer=new StringTokenizer("");
		
		String next() {
			while (!tokenizer.hasMoreTokens()) {
				try {
					tokenizer=new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return tokenizer.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] nextIntArr(int numberOfElements) {
			int[] values=new int[numberOfElements];
			for (int i=0; i<numberOfElements; i++) values[i]=nextInt();
			return values;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}
