Problem:
Disjoint Set

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.

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.

Constraints

1 ≤ n ≤ 10000

1 ≤ q ≤ 100000

x ≠ y

Sample 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

Sample Output

0
0
1
1
1
0
1
1

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
using namespace std;
typedef long long ll;

class UnionFind {
    vector<int> par;
    int c;

  public:
    UnionFind(int n) : par(n, -1), c(n) {}
    int root(int x) { return par[x] < 0 ? x : par[x] = root(par[x]); }
    int size(int x) { return -par[root(x)]; }
    int count() const { return c; }
    bool same(int x, int y) { return root(x) == root(y); }
    bool unite(int x, int y) {
        if((x = root(x)) == (y = root(y))) return false;
        if(par[x] < par[y]) swap(x, y);
        par[y] += par[x];
        par[x] = y;
        return c--;
    }
};

int main() {
    ll n, q, p, a, b;
    cin >> n >> q;

    UnionFind tree(n);

    while(q--) {
        cin >> p >> a >> b;
        if(!p)
            tree.unite(a, b);
        else
            cout << tree.same(a, b) << "\n";
    }
}