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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;



public class A {
	
	
    public static void main(String[] args) {
    
    	FastScanner scan = new FastScanner();
    	                                                                     
    	
    	int t = 1;//scan.nextInt();
    	for(int tt=0; tt<t; tt++) {
    		int a = scan.nextInt();
    		if(a % 2 == 1) {
    			System.out.println(0);
    		} else {
    			long ans = a/2;
    			System.out.println((long)Math.pow(2, ans));
    		}
    	}
    }
    
    
   
    public static void sort(int [] a) {
    	ArrayList<Integer> b = new ArrayList<>();
    	for(int i: a) b.add(i);
    	Collections.sort(b);
    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
    }
    
  
    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();
    	}
    	
    	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 nextLong() {
    		return Long.parseLong(next());
    	}
    	
    	
    }
    
   
}

