Problem: 

Input

The input contains a single integer a (0 ≤ a ≤ 35).

Output

Output a single integer.

Examples

Input

3


Output

8


Input

10


Output

1024

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;


public class A {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		long[]a = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8092, 16184, 32368, 64736, 129472, 258944, 517888, 1035776, 2071552, 4143104, 8286208, 16572416, 33144832, 66289664, 132579328, 265158656, 530317312, 1060634624, 2121269248, 4242538496l, 2*4242538496l, 4*4242538496l, 8*4242538496l};
		System.out.println(a[nextInt()]);
		pw.close();
	}
	private static int divs(int n) {
		int res = 0;
		for (int i = 1; i*i <= n; i++) {
			if (n % i==0) {
				res++;
				if (i*i != n)
					res++;
			}
		}
		return res;
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}