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;

int par[10000], d[10000];

int find(int x) {
	if (x == par[x])
		return x;
	else
		find(par[x]);
}

int main()
{
	int n, q;

	cin >> n >> q;

	for (int i = 0; i < n; i++) {
		par[i] = i;
		d[i] = 1;
	}

	for (int i = 0; i < q; i++) {
		int k, x, y;
		cin >> k >> x >> y;

		if (k == 0) {
			x = find(x);
			y = find(y);
			if (d[x] > d[y])
				par[y] = x;
			else if (d[x] < d[y])
				par[x] = y;
			else {
				par[y] = x;
				d[x]++;
			}
		}
		else if (k == 1) {
			if (find(x) == find(y))
				cout << 1 << endl;
			else
				cout << 0 << endl;
		}
	}

	return 0;
}