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:
/* ---- I.T.N.O.G ----  */
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define ll long long

const ll MAXlog = 60, MAX = 1e4+1, inf = 1e18+1, mod = 1e9+7;

ll x, y, q,f, n, p[MAX];

ll fnd (int x) {
	return ( p[x] == x ? x : p[x] = fnd(p[x]) );
}

void unt ( int x , int y ) {
	p[fnd(y)] = fnd(x);
}

bool is (int x, int y) {
	return fnd(y) == fnd(x);
}

int main() {
	
	ios_base :: sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> q;
	for (int i = 0; i < n; i++) p[i] = i;
	while (q--) {
		cin >> f >> x >> y;
		if (!f)
			unt ( x , y );
		else
			cout << (is (x, y) ? 1 : 0) << endl;
	}
	return 0;


}