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 <stdio.h>
int djs[10010];
int height[10010];
int djsUnion(int x,int y){
	if(height[x]>height[y]){
		djs[y]=x;
	}
	else if(height[x]<height[y]){
		djs[x]=y;
	}
	else{
		djs[x]=y;
		height[y]++;
	}
	return 0;
}
int find (int x){
	if(djs[x]==-1)return x;
	else{
		djs[x]=find(djs[x]);
		return djs[x];
	}
}
int main(){
	int i,n,m,x,y,z;
	scanf("%d %d",&n,&m);
	for(i=0;i<n;i++){
		djs[i]=-1;
		height[i]=1;
	}
	for(i=0;i<m;i++){
		scanf("%d %d %d",&x,&y,&z);
		if(x==1){
			if(find(y)==find(z)){
				printf("1\n");
			}
			else{
				printf("0\n");
			}
		}
		else{
			if(find(y)!=find(z)){
				djsUnion(find(y),find(z));
			}
		}	
	}
	return 0;
}