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<algorithm>
#include<unordered_map>
using namespace std;
int main(){
  int n,q; scanf("%d %d",&n,&q);
  unordered_map<int, int> v;
  for(int i=0;i<n;i++) v[i]=i;
  for(int i=0;i<q;i++){
    int com,x,y; scanf("%d %d %d",&com,&x,&y);
    for(;;)
      if(x==v[x]) break;
      else x=v[x];
    for(;;)
      if(y==v[y]) break;
      else y=v[y];
    if(com){
      printf("%d\n", v[x]==v[y]?1:0);
    } else{
      if(x>y) swap(x,y);
      if(v[x]>v[y])
	v[x]=v[y];
      else{
	int tmp=x;
	for(;;){
	  if(tmp==v[tmp]){
	    v[y]=tmp;
	    break;
	  }
	  else{
	    tmp=v[tmp];
	  }
	}
      }
    }
  }
}