Problem:
Score : 400 points

Problem Statement

There are N cities. There are also K roads and L railways, extending between the cities.
The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities.
No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.

We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads.
We will also define connectivity by railways similarly.

For each city, find the number of the cities connected to that city by both roads and railways.

Constraints

2 ≦ N ≦ 2*10^5

1 ≦ K, L≦ 10^5

1 ≦ p_i, q_i, r_i, s_i ≦ N

p_i < q_i

r_i < s_i

When i ≠ j, (p_i, q_i) ≠ (p_j, q_j)

When i ≠ j, (r_i, s_i) ≠ (r_j, s_j)

Input

The input is given from Standard Input in the following format:

N K L
p_1 q_1
:
p_K q_K
r_1 s_1
:
r_L s_L

Output

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

Sample Input 1

4 3 1
1 2
2 3
3 4
2 3

Sample Output 1

1 2 2 1

All the four cities are connected to each other by roads.

By railways, only the second and third cities are connected. Thus, the answers for the cities are 1, 2, 2 and 1, respectively.

Sample Input 2

4 2 2
1 2
2 3
1 4
2 3

Sample Output 2

1 2 2 1

Sample Input 3

7 4 4
1 2
2 3
2 5
6 7
3 5
4 5
3 4
6 7

Sample Output 3

1 1 2 1 2 2 2

Solution in Ruby (2.3.3) (judge verdict: Accepted)

class Node
  attr_accessor :parent, :rank

  def initialize(n)
    @parent = n
    @rank = 0
  end
end

class UnionFindTree
  # 0 から n まで（計 n+1個）のNodeを作る
  def initialize(n)
    @nodes = (0..n).to_a.map { |i| Node.new(i) }
  end

  # x要素のルート親を返す
  def find(x)
    return x if @nodes[x].parent == x

    @nodes[x].parent = find(@nodes[x].parent)
  end

  # a,b を同一グループ化する
  def unite(a, b)
    a = find(a)
    b = find(b)
    return if a == b

    if @nodes[a].rank < @nodes[b].rank
      @nodes[a].parent = b
    else
      @nodes[b].parent = a
      @nodes[a].rank += 1 if @nodes[a].rank == @nodes[b].rank
    end
  end

  # a,b が同一グループ？
  def same?(a, b)
    find(a) == find(b)
  end
end

def main
  c_num, p_num, r_num = gets.split.map(&:to_i)

  # 道路で行ける都市をグループ化
  path_tree = UnionFindTree.new(c_num)
  p_num.times do
    path_tree.unite(*gets.split.map(&:to_i))
  end

  # 鉄道で行ける都市をグループ化
  rail_tree = UnionFindTree.new(c_num)
  r_num.times do
    rail_tree.unite(*gets.split.map(&:to_i))
  end

  # 同じ親セットを持つ都市をカウントアップしていく
  groups = Hash.new(0)
  1.upto(c_num) do |i|
    pp = path_tree.find(i)
    rp = rail_tree.find(i)
    groups[[pp, rp]] += 1
  end

  # 同じ親セットをもつ都市数を出力
  nums = (1..c_num).inject([]) do |s, i|
    pp = path_tree.find(i)
    rp = rail_tree.find(i)
    s << groups[[pp, rp]]
  end
  puts nums.join(" ")
end

if __FILE__ == $0
  main
end