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<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;

int par[10000], high[10000];
int find(int x) {
	if (par[x] == x)return x;
	return par[x] = find(par[x]);
}
void unite(int x, int y) {
	x = find(x); y = find(y);
	if (x == y)return;
	if (high[x] > high[y])par[y] = x;
	else {
		par[x] = y;
		if (high[x] == high[y])high[y]++;
	}
}
bool same(int x, int y) {
	return find(x) == find(y);
}
int main() {
	int n, q; scanf("%d%d", &n, &q);
	rep(i, n)par[i] = i;
	rep(i, q) {
		int com, x, y; scanf("%d%d%d", &com, &x, &y);
		if (com)printf("%d\n", same(x, y));
		else unite(x, y);
	}
}