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);
    }
}

#include <stdio.h>

int main(){
    long long int n,m=1;
    scanf("%lld",&n);
    while(n>0){
        if(n%2==0)m*=2;
        n/=2;
    }
    printf("%lld\n",m);
    return 0;
}

#include <stdio.h>

int main(){
    long long int n,m=1;
    scanf("%lld",&n);
    while(n>0){
        if(n%2==0)m*=2;
        n/=2;
    }
    printf("%lld\n",m);
    return 0;
}