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;

const int MAX = 10000;

int parent(int uft[], int child) {
	if (uft[child] == -1)return child;
	return uft[child] = parent(uft, uft[child]);
}

int main() {
	int uft[MAX];
	int n;
	int q;

	cin >> n >> q;

	for (int i = 0; i < n; i++)uft[i] = -1;
	for (int i = 0; i < q; i++) {
		int com, x, y;
		cin >> com >> x >> y;
		
		if (com == 0) {
			int xp = parent(uft, x);
			int yp = parent(uft, y);
			if (xp != yp) {
				uft[xp] = yp;
			}
		}
		else {
			if (parent(uft, x) == parent(uft, y)) {
				cout << 1 << endl;
			}
			else {
				cout << 0 << endl;
			}
		}
	}

	return 0;
}