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<cstdio>

int n,q;
int par[10000];
int rank[10000];

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

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;
  if(rank[x]<rank[y]){
    par[x] = y;
  }else{
    par[y] = x;
    if(rank[x]==rank[y]) rank[x]++;
  }
}

void same(int x,int y){
  if(find(x)==find(y)) printf("1\n");
  else printf("0\n");
}

int main(){
  scanf("%d %d",&n,&q);
  init();
  for(int i=0;i<q;i++){
    int cmd,x,y;
    scanf("%d %d %d",&cmd,&x,&y);
    if(cmd==0) unite(x,y);
    if(cmd==1) same(x,y);
  }
  return 0;
}