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 UF {
	int par[100000];
	UF() { for (int i = 0; i < 100000; i++) par[i] = i; }
	int root(int x) { if (x == par[x]) return x; return par[x] = root(par[x]); }
	void unit(int x, int y) { x = root(x); y = root(y); par[x] = y; }
	bool same(int x, int y) { return root(x) == root(y); }
};

int n, q;
int type, x, y;
UF uf;

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