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>
#include<vector>
#include<algorithm>


using namespace std;

int P[10010], rk[10010];

void init(int n) {
	for (int i = 0; i < n; i++) {
		P[i] = i;
		rk[i] = 0;
	}
}

int root(int x) {
	if (P[x] == x)return x;
	return P[x] = root(P[x]);
}

void unite(int x, int y) {
	x = root(x);
	y = root(y);
	if (x == y)return;

	if (rk[x] < rk[y])P[x] = y;
	else {
		P[y] = x;
		if (rk[x] == rk[y])rk[x]++;
	}
}

bool same(int x, int y) {
	return root(x) == root(y);

}

int main(void) {

	int n, q;
	cin >> n >> q;
	init(n);
	while (q--) {
		int com, x, y;
		cin >> com >> x >> y;
		if (com)cout << same(x, y) << endl;
		else unite(x, y);
	}
	return 0;
}