Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1, p2, ..., pn.

Petya decided to introduce the sum operation on the set of permutations of length n. Let's assume that we are given two permutations of length n: a1, a2, ..., an and b1, b2, ..., bn. Petya calls the sum of permutations a and b such permutation c of length n, where ci = ((ai - 1 + bi - 1) mod n) + 1 (1 ≤ i ≤ n).

Operation <image> means taking the remainder after dividing number x by number y.

Obviously, not for all permutations a and b exists permutation c that is sum of a and b. That's why Petya got sad and asked you to do the following: given n, count the number of such pairs of permutations a and b of length n, that exists permutation c that is sum of a and b. The pair of permutations x, y (x ≠ y) and the pair of permutations y, x are considered distinct pairs.

As the answer can be rather large, print the remainder after dividing it by 1000000007 (109 + 7).

Input

The single line contains integer n (1 ≤ n ≤ 16).

Output

In the single line print a single non-negative integer — the number of such pairs of permutations a and b, that exists permutation c that is sum of a and b, modulo 1000000007 (109 + 7).

Examples

Input

3


Output

18


Input

5


Output

1800
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.in;
import static java.lang.System.out;

public class SolutionD extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                                            InputStreamReader(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 parseInt(next());
        }

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

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

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

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        solve();
        out.close();
    }

    static final long MODULO = 1_000_000_007;

    static final long[] result = {1, 3, 15, 133, 2025, 37851, 1030367, 36362925};

    static final long[] f = precalcFactorials(15, MODULO);

    private static long[] precalcFactorials(int maxN, long mod) {
        long[] factorials = new long[maxN + 1];
        factorials[0] = 1;
        for (int i = 1; i < factorials.length; i++) {
            factorials[i] = (factorials[i-1] * i) % mod;
        }
        return factorials;
    }

    private static void solve() {
        int n = scanner.nextInt();
        if (n % 2 == 0) {
            out.println(0);
        } else {
            out.println((result[n/2] * f[n]) % MODULO);
        }
    }
}