Problem Description
Given an positive integer A (1 <= A <= 100), output the lowest bit of A.

For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2.

Another example goes like this: given A = 88, we can write A in binary form as 1011000, so the lowest bit of A is 1000, so the output should be 8.

Input
Each line of input contains only an integer A (1 <= A <= 100). A line containing “0” indicates the end of input, and this line is not a part of the input data.

Output
For each A in the input, output a line containing only its lowest bit.

Sample Input
26
88
0

Sample Output
2
8

输入一个十进制的数：然后转换为二进制，输出二进制的位数从后面到前面数的第一个1的时候的数的十进制。
例如：26二进制是11010，所以从后面对前面数，第二位为1，所以取值为10，再转换成十进制，输出为2；

import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            if(n==0){
                return ;
            }
            String str = Integer.toString(n, 2);
            //System.out.println(str);
            for(int i=str.length()-1;i>=0;i--){
                if(str.charAt(i)=='1'){
                    System.out.println((int)Math.pow(2, str.length()-1-i));
                    break;
                }
            }
        }
    }

}