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<stdio.h>
#include<algorithm>
using namespace std;
int par[100005],com[100005],x[100005],y[100005];
int n,q;

void init(int u){
  for(int i=0;i<u;i++){
    par[i]=i;
  }
}

int find(int x){
  if(par[x]==x){
    return x;
  }else{
    return par[x]=find(par[x]);
  }
}

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

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

int main(){
  scanf("%d%d",&n,&q);
  for(int i=0;i<q;i++){
    scanf("%d%d%d",&com[i],&x[i],&y[i]);
  }

  init(n);

  for(int i=0;i<q;i++){
    if(com[i]==0){
      unite(x[i],y[i]);
    }else{
      printf("%d\n",same(x[i],y[i]));
    }
  }

  return 0;
}