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>

using namespace std;

struct Unionfind{
	vector<int> p;
	Unionfind(int n ):p(n, -1) {}
	int find(int x) {
		if (p[x] < 0) return x;
		return p[x] = find(p[x]);
	}

	bool unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y) return false;
		if (p[x] > p[y]) swap(x, y);
		p[x] += p[y];
		p[y] = x;
		return true;
	}

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

int main() {

	int n, q;;
	cin >> n >> q;
	int c, x, y;
	Unionfind uf(n);
	for (int i = 0; i < q; i++) {
		cin >> c >> x >> y;
		if (c == 0) uf.unite(x, y);
		else {
			cout << uf.same(x, y) << endl;
		}
	}
	return 0;
}