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)

N,R,T = gets.split.map(&:to_i)

ns  = (1..N).to_a
res = (0...R).map{ f,t = gets.split.map(&:to_i); [f,t] }
tes = (0...T).map{ f,t = gets.split.map(&:to_i); [f,t] }

class UnionFind_PathCompression
  attr_reader :rank, :parent, :value
  attr_writer :rank, :parent

  def initialize(x)
    @value = x
    @rank = 0
    @parent = self
  end

  def ==(obj)
    @value == obj.value
  end

  def find
    return self if self.parent == self
    root = self.parent.find
    self.parent = root
    root
  end

  def union(y)
    xr = find; yr = y.find
    return if xr == yr
    if xr.rank >= yr.rank
      yr.parent = xr
      xr.rank += 1
    else
      y.union(self)
    end
  end
end

def minimum_spanning_tree_kruskal(nodes, edges)
  # edges: [[node_from, node_to, weight]]
  edges.sort_by!{ |e| e[2] }
  th = Hash[nodes.map{ |x| [x, UnionFind_PathCompression.new(x)] }]
  i = 0
  while i < edges.size
    e = edges[i]
    i += 1
    f,t,w = *e
    tf = th[f]; tt = th[t]
    next if (tf.find == tt.find)
    tf.union(tt)
  end
  th
end

th = Hash[ns.map{ |n| [n,{ }] }]
minimum_spanning_tree_kruskal(ns, res).each{ |x,uf| th[x][:road] = uf.find.value }
minimum_spanning_tree_kruskal(ns, tes).each{ |x,uf| th[x][:train] = uf.find.value }

ch = {}
th.each{ |x,h| t = h[:road]*N+h[:train]; ch[t] = (ch[t] || 0) + 1 }
th.keys.sort.each{ |n| h = th[n]; puts ch[h[:road]*N+h[:train]] }