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 Ruby (judge verdict: Accepted)

class UnionFind
  def initialize(n)
    @size = Array.new(n, 1)
    @rank = Array.new(n, 0)
    @parent = []

    (0..n).each do |i|
      @parent[i] = i
    end
  end

  def find(x)
    if @parent[x] == x
      x
    else
      @parent[x] = find(@parent[x])
    end
  end

  def unite(x, y)
    x = find(x)
    y = find(y)
    return if x == y

    if @rank[x] < @rank[y]
      @parent[x] = y
      @size[y] += @size[x]
    else
      @parent[y] = x
      @size[x] += @size[y]

      @rank[x] += 1 if @rank[x] == @rank[y]
    end
  end

  def same?(x, y)
    find(x) == find(y)
  end

  def size(x)
    @size[find(x)]
  end
end

N, Q = gets.split.map(&:to_i)
uf = UnionFind.new(N + 1)

Q.times do
  com, x, y = gets.split.map(&:to_i)

  if com == 0
    uf.unite(x, y)
  else
    if uf.same?(x, y)
      puts 1
    else
      puts 0
    end
  end
end