You have a given integer n. Find the number of ways to fill all 3 × n tiles with the shape described in the picture below. Upon filling, no empty spaces are allowed. Shapes cannot overlap.

<image> This picture describes when n = 4. The left one is the shape and the right one is 3 × n tiles. 

Input

The only line contains one integer n (1 ≤ n ≤ 60) — the length.

Output

Print the number of ways to fill.

Examples

Input


4


Output


4

Input


1


Output


0

Note

In the first example, there are 4 possible cases of filling.

In the second example, you cannot fill the shapes in 3 × 1 tiles.
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;

public class SolutionA {
    private static FastScanner in = new FastScanner();
    private static PrintWriter out =
        new PrintWriter(new BufferedOutputStream(System.out));

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

    static void solve() {
        int n = in.nextInt();
        if (n % 2 != 0) {
            out.println(0);
        }
        else {
            out.println(1 << (n / 2));
        }
    }
    
    static void tests() {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    private static class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;
    
        public FastScanner() {
            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;
        }
    }
}
