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<iostream>
#include<vector>
#include<cstdio>
using namespace std;
int p[10001];

int root(int x){
  vector <int> v;
  while(p[x]!=x){
    v.push_back(x);
    x=p[x];
  }
  for(int i=0;i<v.size();i++)p[v[i]]=x;
  return x;
}

void unite(int x,int y){
  x=root(x);
  y=root(y);
  if(x!=y){
    p[y]=x;
  }
}

bool same(int x,int y){
  x=root(x);
  y=root(y);
  if(x==y)return true;
  else return false;
}

int main(){
  int n,q,com,x,y;
  for(int i=0;i<10001;i++)p[i]=i;
  cin>>n>>q;
  for(int i=0;i<q;i++){
    scanf("%d %d %d",&com,&x,&y);
    if(com==0){
      unite(x,y);
    }else{
      if(same(x,y))printf("%d\n",1);
      else printf("%d\n",0);
    }
  }
  return 0;
}