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<iostream>
using namespace std;
const int maxp=10005;
int n,q,Tree[maxp];

int GetP(int s) {
	if(Tree[s]==s) return s;
	else return Tree[s]=GetP(Tree[s]);
}

void Merge(int x,int y) {
	int px=GetP(x),py=GetP(y);
	if(px!=py) Tree[py]=px;
}

bool Check(int x,int y) {
	int px=GetP(x),py=GetP(y);
	if(px==py) return true;
	else return false;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cin.tie(NULL);
	cin>>n>>q;
	for(int i=0; i<n; i++)
		Tree[i]=i;
	int judge,x,y;
	while(q--) {
		cin>>judge>>x>>y;
		if(judge) {
			if(Check(x,y)) cout<<"1"<<endl;
			else cout<<"0"<<endl;
		} else Merge(x,y);
	}
	return 0;
}