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 ll long long
#define endl '\n'

const int maxn = 10005;
vector< int > v(maxn);

int find(int a){
	if(a == v[a]) return a;
	return v[a] = find(v[a]);
}

void merge(int a, int b){
	a = find(a), b = find(b);
	if(a == b) return;
	v[b] = a;
}

void solve(){
	int n, q; cin>>n>>q;
	for(int i = 1; i<=n; i++) v[i] = i;
	for(int i = 0; i<q; i++){
		int a, x, y; cin>>a>>x>>y;
		if(a){
			cout<<(find(x) == find(y))<<endl;
		}else{
			merge(x, y);
		}
	}
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t1 = 1; //cin>>t1;
	while(t1--){
		solve();
	}
	return 0;
}