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. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

// Compete Against yourself :)

import java.io.*;
import java.util.*;

public class Task1391C {
    static FastReader fr = new FastReader();
    StringBuilder op = new StringBuilder();
    final long mod = 1_000_000_007;
    public void run() {
        long n = fr.nextLong();
        long facn = 1;
        for(int i=1;i<=n;i++){
            facn = (facn*i)%mod;
        }
        long answer = facn - power(2,n-1);
        if(answer<0){
            answer+=mod;
        }
        System.out.println(answer);
    }

    public static void main(String[] args) {
        new Task1391C().run();
    }
    //math related functions

    //gcd
    private long gcd(long a, long b) {
        if (a == 0) {
            return b;
        } else {
            return gcd(b % a, a);
        }
    }

    //lcm
    private long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    //exponential power
    private long power(long a, long b) {
        if (a == 1 || b == 0) {
            return 1;
        }
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                res = (res * a)%mod;
            }
            b /= 2;
            a = (a*a)%mod;
        }
        return res%mod;
    }

    // euler totient function copied from
    // https://www.geeksforgeeks.org/eulers-totient-function/

    private int phi(int n) {
        // Initialize result as n
        double result = n;

        // Consider all prime factors of n and for
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 2; p * p <= n; ++p) {
            // Check if p is a prime factor.
            if (n % p == 0) {
                // If yes, then update n and result
                while (n % p == 0)
                    n /= p;
                result *= (1.0 - (1.0 / (double) p));
            }
        }

        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (n > 1)
            result *= (1.0 - (1.0 / (double) n));

        return (int) result;
    }


    //binary search related functions
    private int lowerBound(int[] a, int key) {
        int ans = -1;
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = (low + high) >>> 1;
            if (a[mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    private int upperBound(int[] a, int key) {
        int ans = -1;
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = (low + high) >>> 1;
            if (a[mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    //reverse
    private void reverse(int[] a, int s, int e) {
        for (int i = s, j = e; i <= j; i++, j--) {
            int temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }

    private void reverse(long[] a, int s, int e) {
        for (int i = s, j = e; i <= j; i++, j--) {
            long temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }

    //sorting
    private void sort(int[] a) {
        var ls = new ArrayList<Integer>();
        for (int x : a) {
            ls.add(x);
        }
        Collections.sort(ls);
        for (int i = 0; i < a.length; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sort(long[] a) {
        var ls = new ArrayList<Long>();
        for (long x : a) {
            ls.add(x);
        }
        Collections.sort(ls);
        for (int i = 0; i < a.length; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sortInRangeInt(int[] a, int s, int e) {
        List<Integer> ls = new ArrayList<>();
        for (int i = s; i <= e; i++) ls.add(a[i]);
        Collections.sort(ls);
        for (int i = s; i <= e; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sortInRangeLong(long[] a, int s, int e) {
        List<Long> ls = new ArrayList<>();
        for (int i = s; i <= e; i++) ls.add(a[i]);
        Collections.sort(ls);
        for (int i = s; i <= e; i++) {
            a[i] = ls.get(i);
        }
    }

    //Fast I/O
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException ie) {
                    ie.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        ;

        int[] readIntArray(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;
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException ie) {
                ie.printStackTrace();
            }
            return str;
        }
    }
}
