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>

using namespace std;

class UFT{
public:
  int P[10010];//その頂点の親を格納する

  //初めは全ての頂点はバラバラで違うグループに属する
  void init(int N){
    for(int i=0;i<N;i++) P[i]=i;
  }

  int root(int a){
    if(P[a]==a) return a;
    P[a]=root(P[a]);//aの親を辿っていく
    return P[a];
  }

  bool is_same_set(int a,int b){
    return root(a)==root(b);
  }

  void unite(int a,int b){
    //aのルートだったもののルートをbのルートにする
    P[root(a)]=root(b);
  }
};

int main(){
  int N,Q;cin >> N >> Q;
  int com,x,y;
  UFT P;P.init(N);
  for(int i=0;i<Q;i++){
    cin >> com >> x >> y;
    if(com==0){
      P.unite(x,y);
    }else{
      int t=(int)(P.is_same_set(x,y));
      cout << t << endl;
    }
  }
}