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 par[10010];

int find(int x){
    if(par[x] == x){
        return x;
    }else{
        return par[x] = find(par[x]);  //経路圧縮
    }
}

int same(int x,int y){
   return (find(x) == find(y));
}

void unite(int x,int y){
    x = find(x);
    y = find(y);
    if(x == y) return;
    par[x] = y;
}

int main(){
    int N,Q;
    cin >> N >> Q;
    
    //init
    for(int i = 0;i < N;i++) par[i] = i;
    //par[i] = i ならば根
    for(int i = 0;i < Q;i++){
        int com,x,y;
        cin >> com >> x >> y;
        if(com){
            cout << same(x,y) << endl;
        }else{
            unite(x,y);
        }
    }
    
    
}