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.IOException;
import java.util.Scanner;

public class pb3 {

	static final int MOD = 1000000007;

	static int mmod(long a, int b) {
		return (int) ((a * b) % MOD);
	}

	static int amod(int a, int b) {
		return (a + b) % MOD;
	}

	static int smod(int a, int b) {
		return (a - b + MOD) % MOD;
	}

	public static void main(String[] args) throws IOException {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();
		in.close();
		memoFact[0] = 1; memoFact[1] = 1;
		int tot = modFactMemo(n);
		for (int i = 0; i < n/2; i++) {
			tot = smod(tot, mmod(modComb(n-1, i), 2));
		}
		if (n%2 == 1) tot = smod(tot, modComb(n-1, n/2));
		System.out.println(tot);
		
	}
	
	static int[] memoFact = new int[1000001]; // initialize memoFact[0, 1] to 1
	static int maxFact = 1;
	static int modFactMemo(int a) {
		if (a <= maxFact) return memoFact[a];
		for (; maxFact < a; maxFact++) {
			memoFact[maxFact+1] = mmod(memoFact[maxFact], maxFact+1);
		}
		return memoFact[a];
	}
	
	static int modComb(int n, int k) {
		if (n < 0 || k < 0 || k > n) return 0;
		if (k == 0 || k == n) return 1;
		return mmod(mmod(modFactMemo(n), modInv(modFactMemo(k))), modInv(modFactMemo(n-k)));
	}
	
	// e cannot be negative
	static int fastModPow(int b, int e) {
		int res = (e & 1) > 0 ? b : 1;
		while (e != 0) {
			e >>= 1;
			b = mmod(b, b);
			if ((e & 1) > 0) res = mmod(res, b);
		}
		return res;
	}
	
	// a should not be 0 (would return 0)
	static int modInv(int a) {
		return fastModPow(a, MOD-2);
	}
}
