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;

int find(vector<int> &C, int x){ return (C[x] == x) ? x : find(C, C[x]);}
bool same(vector<int> &C, int x, int y){ return find(C, x) == find(C, y);}
void merge(vector<int> &C, int x, int y){ C[find(C,x)] = find(C, y);}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  #ifdef Larra
  freopen("in","r",stdin);
  freopen("out","w",stdout);
  #endif
  int n, q;
  cin>>n>>q;
  vector<int> C(n);
  for(int i = 0; i < n; i++) C[i] = i;
  for(int i = 0; i < q; i++){
    int com, x, y;
    cin>>com>>x>>y;
    if(com)
      cout<<same(C, x,y)<<"\n";
    else
      merge(C, x,y);
  }
}