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;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
int par[111111],ran[111111];
void init(int n){
	rep(i,n){
		par[i]=i;
		ran[i]=0;
	}
}
int find(int x){
	if(par[x]==x)return x;
	return par[x]=find(par[x]);
}
void unite(int x,int y){
	x=find(x),y=find(y);
	if(x==y)return;
	if(ran[x]<ran[y])par[x]=y;
	else{
		par[y]=x;
		if(ran[x]==ran[y])ran[x]++;
	}
}
bool same(int x,int y){
	return find(x)==find(y);
}
int n,q,a,b,c;
signed main(){
	cin>>n>>q;
	init(n);
	while(q--){
		cin>>a>>b>>c;
		if(a)cout<<same(b,c)<<endl;
		else unite(b,c);
	}
	return 0;
}