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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;                }            }        }    }}

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