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>
#define MAX 10001
using namespace std;
int ata[MAX], n, q, a, b, c;

void build(){
    for(int i=0;MAX>i;i++){
        ata[i]=i;
    }
}

int find_set(int v){
    if(v==ata[v]){
        return v;
    }
    return ata[v]=find_set(ata[v]);
}

void union_set(int b, int c){
    b=find_set(b);  c=find_set(c);
    if(b!=c){
        ata[c]=b;
    }
}

int main(){
    ios::sync_with_stdio(false);    cin.tie(NULL);  cout.tie(NULL);
    build();    cin >> n >> q;
    for(int i=0;q>i;i++){
        cin >> a >> b >> c;
        if(a){
            cout << (find_set(b)!=find_set(c) ? 0 : 1) << "\n";     continue;
        }
        union_set(b, c);
    }
}