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 <bits/stdc++.h>
using namespace std;
vector<int> data(10001,-1);

int root(int x){
  if(data[x] == -1)return x;
  else return data[x] = root(data[x]);
}

bool same(int x,int y){
  if(root(x) == root(y))return 1;
  else return 0;
}

int unite(int x,int y){
  if(!same(x,y))data[root(x)] = y;
  return 0;
}

int main(){
  int n,q,com,x,y;
  vector<int> ans;
  cin >> n >> q;
  while(data.size()>n+1)data.erase(data.begin());
  for(int i = 0;i < q;i++){
    cin >> com >> x >> y;
    if(com){
      if(same(x,y))ans.push_back(1);
      else ans.push_back(0);
    }
    else{
      unite(x,y);
    }
  }
  for(int i = 0;i < ans.size();i++)cout << ans[i] << endl;
}