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<bits/stdc++.h>
using namespace std;
int main(void){
	int n,q;
	int m,a,b,x,y;
	int par[10000],ran[10000];
	int i;
	cin>>n>>q;
	for(i=0;i<n;i++)	par[i]=i,ran[i]=0;
	for(i=0;i<q;i++){
		cin>>m>>a>>b;
		x=a;	y=b;
		while(x!=par[x])	x=par[x];
		while(y!=par[y])	y=par[y];
		if(m==0){
			if(x!=y){
				if		(ran[x]>ran[y])	par[y]=x;
				else if	(ran[x]<ran[y])	par[x]=y;
				else					par[y]=x,ran[x]++;
			}
		}
		else{
			if(x==y)	cout<<1<<endl;
			else		cout<<0<<endl;
		}
	}
	return 0;
}