public static long sumXor(long n) {
// Write your code here
Long res = 0L, nn = n;
res = (long)Math.pow(2,(Long.toBinaryString(nn).chars().filter(x -> x == '0').count()));
return n == 0 ? 1 : res;
}

fn sumXor(n: i64) -> i64 {
    let zeros = n.count_zeros() - n.leading_zeros();
    2_i64.pow(zeros)
}

def sumXor(n):
    # So basically, criteria is met when all bits are flipped to 1's for an n xor x 
    #and none are flipped to zeros
    #Meaning any non-1 in n is an option for x values, fortunately x is never greater than n making
    #this a lot easier
    if n==0:return 1
    return 2**sum([True if i=='0' else False for i in format(n,'0b')])

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        long count = 0;
        while(n != 0){
            count += (n%2 == 0)?1:0;
            n/=2; 
        }
        count = (long) Math.pow(2,count);
        System.out.println(count);
    }
}