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 <iostream>
using namespace std;

int par[10001], depth[10001];

void init(int n){
 
  for(int i = 0; i < n; i++) par[i] = i, depth[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(depth[x] < depth[y]) par[x] = y;
  else{
    par[y] = x;
    if(depth[x] == depth[y]) depth[x]++;
  }

}

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

int main(){

  int n, q;
  cin >> n >> q;
  init(n);
  
  while(q--){
    int com, x, y;
    cin >> com >> x >> y;
    if(!com) unite(x, y);
    else cout << same(x, y) << endl;
  }

}