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 <vector>
#include <iostream>
using namespace std;
int n, q, com, x, y; int group[10000]; vector<int> f[10000];
int main() {
	scanf("%d%d", &n, &q);
	for(int i = 0; i < n; i++) group[i] = i, f[i].push_back(i);
	for(int i = 0; i < q; i++) {
		scanf("%d%d%d", &com, &x, &y);
		if(com == 0) {
			if(group[x] == group[y]) continue;
			x = group[x], y = group[y];
			if(f[y].size() > f[x].size()) swap(x, y);
			for(int j = f[y].size() - 1; j >= 0; j--) {
				group[f[y][j]] = group[x];
				f[x].push_back(f[y][j]);
				f[y].pop_back();
			}
		}
		else printf("%d\n", (group[x] == group[y] ? 1 : 0));
	}
	return 0;
}