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.math.*; 

public class Main {
	static long mod=1_000_000_000+7;
	static long[] facts=new long[2_000_000];
	static long mul(long a, long b) {
		return a*b%mod;
	}
	
	static long add(long a, long b) {
		return (a+b)%mod;
	}
	
	static void precomp() {
		facts[0]=1;
		for (int i=1; i<facts.length; i++) facts[i]=mul(facts[i-1], i);
	}
	
	static long exp(long base, long pow) {
		if (pow==0) return 1;
		long half=exp(base, pow/2);
		if (pow%2==0) return mul(half, half);
		return mul(half, mul(half, base));
	}
	
	static long modInv(long x) {
		return exp(x, mod-2);
	}
	
	static long nCk(int n, int k) {
		return mul(facts[n], mul(modInv(facts[k]), modInv(facts[n-k])));
	}

	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
//		int t=f.nextInt();
		int t=1;
		precomp();
		PrintWriter out=new PrintWriter(System.out);
		while(t>0) {
			t--;
			int n=f.nextInt();
			long ans=facts[n];
			ans=add(ans,-exp(2,n-1));
			if(ans<0) {
				ans+=mod;
			}
			System.out.println(ans);
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new 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());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 