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 P[10010]; // 0 から 10000 までの頂点を取り扱い可能
void init(int N) { // 初期化 はじめは全ての頂点はバラバラ
  for (int i=0; i<N; i++) P[i] = i;
}
int root (int a) { //aの大表現を求める
  if (P[a] == a) return a;
  return (P[a] = root(P[a]));
}
bool is_same_set (int a, int b){
  return root(a) == root(b);
}
void unite(int a, int b){
  P[root(a)] = root(b);
}

int main(){
  int n;
  int q;
  cin >> n >> q;
  int com;
  int x;
  int y;
  init(n + 10);
  int i = 0;
  while  (i < q){
    cin >> com >> x >> y;
    if (com == 0){
      unite (x, y);
    }
    if (com == 1){
      cout << is_same_set(x,y) << endl;
    }
    i++;
  }
  return 0;
}