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>
#include<functional>
#include<algorithm>
using namespace std;
int main(void)
{
	int x,y,n,q,i,j,root[10000],len[10000],c,a,b;
	scanf("%d %d",&n,&q);
	for(i=0;i<n;i++)	{
		root[i]=i;
		len[i]=1;
	}
	for(i=0;i<q;i++)	{
		scanf("%d %d %d",&c,&x,&y);
//		printf("%d %d %d\n",c,x,y);
		while(x!=root[x])	x=root[x];
		while(y!=root[y])	y=root[y];
		if(c==1)	{
			if(x==y)	printf("1\n");
			else printf("0\n");
		}
		else {
			if(x!=y)	{
				if(len[x]>len[y])	{
					root[y]=x;
				}
				else if(len[x]<len[y])	{
					root[x]=y;
				}
				else {
					root[y]=x;
					len[x]++;
				}
			}
		}
	}
	return 0;
}