这个题目就有意思了，要求输入一个数n，然后有一行数，都是0，我们可以设定为n个，然后是经过n次 的转换，转换有n次，假设当前使i，如果n%i==0的话，就变换一次（0变成1，1变成0）


Problem Description


There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).


 


Input


Each test case contains only a number n ( 0< n<= 10^5) in a line.



Output


Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).


 


Sample Input


1 5


 


Sample Output


Hint

hint


Consider the second test case: The initial condition : 0 0 0 0 0 … After the first operation : 1 1 1 1 1 … After the second operation : 1 0 1 0 1 … After the third operation : 1 0 0 0 1 … After the fourth operation : 1 0 0 1 1 … After the fifth operation : 1 0 0 1 0 … The later operations cannot change the condition of the fifth lamp any more. So the answer is 0. 



import java.util.Scanner;

public class P2053 {
  private static Scanner scanner;
  public static void main(String[] args) {
    scanner = new Scanner(System.in);
    while(scanner.hasNext()){
      int n = scanner.nextInt();
      int k = 0;
      for (int i = 1; i <= n; i++) {
        if(n%i == 0){//如果是倍数就改变
          if(k==0){
            k = 1;
          }else {
            k = 0;
          }
        }
      }
      System.out.println(k);
    }
  }
}