Problem: 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.

Constraints

* 1 ≤ n ≤ 10000
* 1 ≤ q ≤ 100000
* x ≠ y

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.

Example

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


Output

0
0
1
1
1
0
1
1

Solution (in CPP):

Solution:
#include<stdio.h>
//http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_1_A
int find(int a,int b,int a1[]){
	return a1[a]==a1[b]?1:0;
}
void union1(int a,int b, int a1[], int n)
{
	int temp = a1[a];//a1[a] may change
	for(int i = 0; i<n; i++){
		if(a1[i]==temp) a1[i] = a1[b];
	}
}


int main(){
	int n,q;
	scanf("%d %d",&n,&q);
	int a1[n+1];//start with 1;
	for(int i = 0; i<n; i++) a1[i] = i;
	for(int i = 0; i<q; i++){
		int a, b;
		int com = 0;
		scanf("%d %d %d",&com,&a,&b);
		if(com==0){
			union1(a,b,a1,n);
			/*for(int i = 0; i<n; i++){
				printf("%d ",a1[i]);
			}printf("\n");*/
		} else {
			printf("%d\n",find(a,b, a1));
		}
	}
	
	
	return 0;
}