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. 
/*
								"जय श्री कृष्णा"
*/
import java.util.*;
import java.io.*;

public class Cyclic_Permutations_C implements Runnable {

	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;
		}

		int[] nextIntArray(int size) {
			int[] arr = new int[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextInt();
			return arr;
		}

		long[] nextLongArray(int size) {
			long[] arr = new long[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextLong();
			return arr;
		}

		double[] nextDoubleArray(int size) {
			double[] arr = new double[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextDouble();
			return arr;
		}
	}

	static int mod = (int) (1e9 + 7);
	static final PrintWriter out = new PrintWriter(System.out);
	static final FastReader fr = new FastReader();

	public static void main(String[] args) throws java.lang.Exception {
		new Thread(null, new Cyclic_Permutations_C(), "Main", 1 << 26).start();
	}
// we have to count total number of non unimodular permutation
//	i.e. n!-2^(n-1) (total permutation - noncyclic permutation)
	void solve() {
		long n=fr.nextLong();
		long res=1;
		for(int i=(int)n; i>=1; i--) {
			res=(res*i)%mod;
		}
		res-=CP.Modular_Expo(2, n-1);
		out.print(res<0?(res+mod)%mod:res);
	}

	@Override
	public void run() {
		long start = System.nanoTime(); // Program Start
		boolean testcase = false;
		int t = testcase ? fr.nextInt() : 1;
		while (t-- > 0) {
			solve();
			out.print("\n");
			out.flush();
		}
		long end = System.nanoTime(); // Program End
		System.err.println("Time taken: " + (end - start) / 1000000 + " ms");
	}

	static class CP {

		static boolean isPrime(long n) {
			if (n <= 1)
				return false;
			if (n == 2 || n == 3)
				return true;
			if (n % 2 == 0 || n % 3 == 0)
				return false;
			for (int i = 5; i * i <= n; i += 6) {
				if (n % i == 0 || n % (i + 2) == 0)
					return false;
			}
			return true;
		}

		static long binary_Expo(long a, long b) { // calculating a^b
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res *= a;
					--b;
				}
				a *= a;
				b /= 2;
			}
			return res;
		}

		static long Modular_Expo(long a, long b) {
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res = (res * a) % mod;
					--b;
				}
				a = (a * a) % mod;
				b /= 2;
			}
			return res;
		}

		static int i_gcd(int a, int b) {// iterative way to calculate gcd.
			while (true) {
				if (b == 0)
					return a;
				int c = a;
				a = b;
				b = c % b;
			}
		}

		static long gcd(long a, long b) {// here b is the remainder
			if (b == 0)
				return a; //because each time b will divide a.
			return gcd(b, a % b);
		}

		static long ceil_div(long a, long b) { // a numerator b denominator
			return (a + b - 1) / b;
		}

		static int getIthBitFromInt(int bits, int i) {
			return (bits >> (i - 1)) & 1;
		}

		static int upper_Bound(int a[], int x) {//closest to the left+1
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] <= x)
					l = m;
				else
					r = m;
			}
			return l + 1;
		}

		static int lower_Bound(int a[], int x) {//closest to the right
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] >= x)
					r = m;
				else
					l = m;
			}
			return r;
		}

		static void sort(int a[]) {
			PriorityQueue<Integer> q = new PriorityQueue<>();
			for (int i = 0; i < a.length; i++)
				q.add(a[i]);
			for (int i = 0; i < a.length; i++)
				a[i] = q.poll();
		}

		static int getMax(int arr[], int n) {
			int mx = arr[0];
			for (int i = 1; i < n; i++)
				if (arr[i] > mx)
					mx = arr[i];
			return mx;
		}
	}

}
