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.*;
import java.util.*;
import java.util.concurrent.TimeUnit;

public class c1391 {
	static long MOD = 1000000007;
    public static void main(String[] args) {
    	FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		
        int N = sc.nextInt();
		
		long[] fact = new long[N+1];
		fact[0] = 1;
		for(int i = 1; i < N+1; i++) {
			fact[i] = (fact[i-1] * i) % MOD;
		}
		
		
		long fail = 0;
		for(int i = 0; i < N/2; i++) {
			fail += 2 * (binpow(fact[i], MOD-2, MOD) * fact[N-1]) % MOD * (binpow(fact[N-1-i], MOD-2, MOD)) % MOD;
			//fail += 2 * (modDivide(modDivide(fact[N-1], fact[i]), fact[N-1-i]) % MOD);
			fail %= MOD;
		}
		if(N % 2 == 1) {
			int i = N/2;
			fail += (binpow(fact[i], MOD-2, MOD) * fact[N-1]) % MOD * (binpow(fact[N-1-i], MOD-2, MOD)) % MOD;
			fail %= MOD;
		}
		out.println((fact[N] - fail + MOD) % MOD);
        
		out.close();
    }
    
    static long binpow(long a, long b, long m) {
		a %= m;
		long res = 1;
		while (b > 0) {
			if ((b & 1) == 1)
				res = res * a % m;
			a = a * a % m;
			b >>= 1;
		}
		return res;
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

		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;
		}
	}

}