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 C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
typedef long long ll;

struct UnionFind {
	vector<int> par;
	vector<int> rank;
	
	UnionFind(int N){
		rep(i,N){
			par.push_back(i);
			rank.push_back(0);
		}
	}
	
	int root(int x) {
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}
	
	void unite(int x, int y){
		int rx = root(x);
		int ry = root(y);
		if(rx == ry) return;
		if(rank[x] < rank[y]){
			par[rx] = ry;
		}else{
			par[ry] = rx;
			if(rank[rx] == rank[ry]) rank[rx]++;
		}
	}
	
	bool same(int x, int y){
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

int main(){
	int N,Q;
	cin >> N >> Q;
	UnionFind tree(N);
	
	rep(i,Q){
		int p,a,b;
		cin >> p >> a >> b;
		if(p == 1){
			cout << tree.same(a,b) << endl;
		}
		if(p == 0){
			tree.unite(a,b);
		}
	}
	
	return 0;
}