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>
using namespace std;

class UnionFindTree {
private:
  int P[10000];
public:
  void init(int N) {
    for (int i=0; i<N; i++) {
      P[i]=i;
    }
  }
  int root(int a) {
    if (P[a]==a) {
      return a;
    } else {
      return (P[a]=root(P[a]));
    }
  }
  void unite(int a,int b) {
    P[root(a)]=root(b);
  }
  bool same(int a,int b) {
    return (root(a)==root(b));
  }
};
int main() {
  int n, q;
  cin >> n >> q;
  UnionFindTree S;
  S.init(n);
  while (q--) {
    int com, x, y;
    cin >> com >> x >> y;
    if (!com) {
      S.unite(x,y);
    } else {
      printf("%d\n",S.same(x,y));
    }
  }
  return 0;
}