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.*;

public class Solver {

    public static void main(String[] args) {
        final long MOD = 1000_000_000 + 7;
        FastReader in = new FastReader();
        // int t = in.nextInt();

        long[] fact = new long[1000_000 + 3];
        long[] pow = new long[1000_000 + 3];

        fact[0] = 1;
        pow[0] = 1;
        for (int i = 1; i < fact.length; i++) {
            fact[i] = ((((long) i % MOD) * fact[i - 1]) % MOD);
            pow[i] = (pow[i-1] * (long) 2) % MOD;
        }

        int n = in.nextInt();
        // long res = ((fact[n - 1] - 1) * (n - 1)) % MOD;
        long res = ((fact[n] - ((long) pow[n-1] % MOD)) + (MOD)) % MOD;
        System.out.println(res);
        // while (t-- > 0) {
        // solve(in);
        // }
    }

    public static void solve(FastReader in) {
        int n = in.nextInt();
    }

    public static void printArray(Object[] a) {
        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println();
    }

    static class Pair {
        public int a, b;

        Pair(Integer a, Integer b) {
            this.a = a;
            this.b = b;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        private 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;
        }
    }

}