Problem:
Disjoint Set

Write a program which manipulates a disjoint set S = {S1, S2, . . . , Sk}.

First of all, the program should read an integer n, then make a disjoint set where each element consists of 0, 1, ... n−1 respectively.

Next, the program should read an integer q and manipulate the set for q queries. There are two kinds of queries for different operations:

unite(x, y): unites sets that contain x and y, say Sx and Sy, into a new set.

same(x, y): determine whether x and y are in the same set.

Input

n q
com1 x1 y1
com2 x2 y2
...
comq xq yq

In the first line, n and q are given. Then, q queries are given where com represents the type of queries. '0' denotes unite and '1' denotes same operation.

Output

For each same operation, print 1 if x and y are in the same set, otherwise 0, in a line.

Constraints

1 ≤ n ≤ 10000

1 ≤ q ≤ 100000

x ≠ y

Sample Input

5 12
0 1 4
0 2 3
1 1 2
1 3 4
1 1 4
1 3 2
0 1 3
1 2 4
1 3 0
0 0 4
1 0 2
1 3 0

Sample Output

0
0
1
1
1
0
1
1

Solution in JAVA (judge verdict: Accepted)

import java.util.Scanner;

class Main {
	static int s[];
	
	public static void main(String args[]) {
		Scanner scan = new Scanner(System.in);
		
		int n, q, com, x, y;
		
		n = scan.nextInt();
		q = scan.nextInt();
		
		s = new int[n];
		
		for(int i = 0; i < n; i++) {
			s[i] = -1;
		}
		
		for(int i = 0; i < q; i++) {
			com = scan.nextInt();
			x = scan.nextInt();
			y = scan.nextInt();

			x = Search(x);
			y = Search(y);
			
			if(com == 0 && x != y) { //unite x??¨y?????\?????????????????????????????????????????¨?????????
				if(y < x) {
					int tmp = x;
					x = y;
					y = tmp;
				}
				
				s[x] += s[y];
				s[y] = x;
				
			} else if(com == 1) {//same
				if(x == y) {
					System.out.println(1);
				} else {
					System.out.println(0);
				}
			}
		}
		
		scan.close();
	}
	
	public static int Search(int x){ 
		//???????????????????????°???????´??????????????????°????????¨???????????\?????????????????????????±???????
		if(s[x] < 0) {
			return x;
		} else {
			return s[x] = Search(s[x]);
		}
	}
}