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 Main {

    static int M = 1_000_000_007;

    private static int testCase(int n) {
        if (n <= 2) return 0;

        long[] factModM = new long[n + 1];
        long[] res = new long[n + 1];

        factModM[0] = 1;
        res[3] = 2;

        for (int i = 1; i <= n; i++) {
            factModM[i] = (factModM[i - 1] * i) % M;
        }

//        for (int i = 0; i <= n; i++) {
//            System.out.print(factModM[i] + " ");
//        }
//        System.out.println();

        //res[n] = 2 * res[n - 1] + (n-1)!(n - 2)
        for (int i = 4; i <= n; i++) {
            res[i] = (((2 * res[i - 1]) % M) +
                    ((factModM[i - 1] * (i - 2)) % M)) % M;
        }

        return (int) res[n];
    }



    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
//        int t = in.nextInt();  // Scanner has functions to read ints, longs, strings, chars, etc.
        //in.nextLine();
//        for (int i = 1; i <= t; ++i) {
        int n = in.nextInt();

        out.println(testCase(n));
//        }

        out.close();
    }

    private 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();
        }

        char[] readCharArray(int n) {
            char[] arr = new char[n];
            try {
                br.read(arr);
                br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return arr;
        }

        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[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}