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: Time Limit Exceeded)

io = STDIN
class UnionFind
  attr_reader :pool
  def initialize(n)
    @node=Array.new(n+1)
    @pool=Hash.new{|h,k|h[k]=[]}
  end
  def make_set(x)#: 未設定の場合、要素がただ１つである新しい集合を作る
    (@node[x]=x;@pool[x]=[x]) unless @node[x]
  end
  def find_set(x)#: 要素が属する集合の"代表"の要素を求める
    make_set(x)
    (@node[x]=find_set(@node[x])) unless @node[x]==x
    @node[x]
  end
  def union(x,y) #: 指定された２つの要素を合併する
    link(find_set(x),find_set(y))
  end
  def link(x,y)
    return if x==y
    a,b=[x,y].sort
    @node[b]=a
    @pool[a].concat(@pool[b])
    @pool[b]=[]
  end
  def same?(a,b)
    find_set(a)==find_set(b)
  end
end
n,k,l=io.gets.chomp.split.map(&:to_i)
@memo={}
road_u=UnionFind.new(n)
rail_u=UnionFind.new(n)
pq,rs=[],[]

k.times do
  pq << io.gets.chomp.split.map(&:to_i)
end

pq.sort.each do |p,q|
  road_u.union(p,q)
end
l.times do
  rs << io.gets.chomp.split.map(&:to_i)
end
rs.sort.each do |r,s|
  rail_u.union(r,s)
end
def calc(i,road_u,rail_u)
  rd=road_u.find_set(i)
  rl=rail_u.find_set(i)
  return @memo[[rd,rl]] if @memo.has_key?([rd,rl])
  if rd == nil || rl == nil
    @memo[[rd,rl]] = 1
  else
    @memo[[rd,rl]] = (road_u.pool[rd] & rail_u.pool[rl]).size
  end
end
ar= (1..n).map do |i|
  calc(i,road_u,rail_u)
end.to_a
puts ar.join(" ")