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 <cstdio>
using namespace std;
const int MAXN = 1e4 + 7;
int p[MAXN], rank[MAXN];
int find(int x) {
	return p[x] == x ? x : (p[x] = find(p[x]));
}
int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++) {
    	p[i] = i;
	}
	while (q--) {
		int op, x, y;
		scanf("%d%d%d", &op, &x, &y);
		if (op) {
			if (find(x) == find(y)) printf("1\n");
			else printf("0\n");
		}
		else {
			int px = find(x), py = find(y);
			if (rank[px] > rank[py]) p[py] = px;
			else if (rank[px] < rank[py]) p[px] = py;
			else {
				p[py] = px;
				rank[px]++;
			}
		}
	}
	return 0;
}