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;

struct Node{
	int b;
	int rank = 0;
};

Node Tree[10000];

int Find(int x) {
	int y = Tree[x].b;
	while (y != Tree[y].b) { y = Tree[y].b; }
	return y;
}

void Unite(int x, int y) {
	int x0 = Find(x), y0 = Find(y);
	Tree[x0].b = y0;
	if (Tree[x0].rank == Tree[y0].rank) Tree[y0].rank++;
	else if (Tree[x0].rank == 0) Tree[x0].rank++;
}

int main() {
	int n, q, com, x, y;
	cin >> n >> q;
	for (int i = 0; i < n; i++) Tree[i].b = i;
	for (int i = 0; i < q; i++) {
		cin >> com >> x >> y;
		if (com == 0) Unite(x, y);
		else if (com == 1) {
			if (Find(x) == Find(y)) cout << 1 << endl;
			else cout << 0 << endl;
		}
	}
}