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:
// be name khoda
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 1e4 + 10;

int n, q;
int par[N];

int root(int v) {
	return (par[v] == -1? v: par[v] = root(par[v]));
}

int main() {
	ios_base :: sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
	cin >> n >> q;
	for (int i = 0; i < n; i++)
		par[i] = -1;
	for (int i = 0; i < q; i++) {
		int c, a, b;
		cin >> c >> a >> b;
		if (c == 0) {
			if (root(a) != root(b))
				par[root(a)] = root(b);
		}
		else {
			if (root(a) == root(b))
				cout << 1 << endl;
			else
				cout << 0 << endl;
		}
	}
	return 0;
}