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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

public class MainClass
{
	public const long Giri = 1000000007;

	public struct Pair
	{
		public int X {get;set;}
		public int Y {get;set;}
	}
	
	public static void Main(string[] args)
	{
		var u = Input().ToInt();
		var n = u[0];
		var k = u[1];
		var l = u[2];
		
		var roads = new UnionFind(n);
		var rails = new UnionFind(n);
		
		for(int i = 0; i < k; i++)
		{
			u = Input().ToInt();
			var x = u[0];
			var y = u[1];
			
			roads.Unite(x-1, y-1);
		}
		
		for(int i = 0; i < l; i++)
		{
			u = Input().ToInt();
			var x = u[0];
			var y = u[1];
			
			rails.Unite(x-1, y-1);
		}
		
		var list = new Dictionary<Pair, int>();
		
		for(int i = 0; i < n; i++)
		{
			var pair = new Pair(){X=roads.Root(i), Y=rails.Root(i)};
			
			if(!list.ContainsKey(pair))
				list.Add(pair, 0);
			list[pair]++;
		}
		for(int i = 0; i < n; i++)
		{
			Console.Write("{0} ", list[new Pair() {X=roads.Root(i), Y=rails.Root(i)}]);
		}
		Console.WriteLine();
	}

	
	static long GCD(long a, long b)
	{
		var v = new[] { a, b };
		while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
		return v[0];
	}
	public static string[] Input() { return Console.ReadLine().Split(' '); }
	public static int Clamp(int val, int max, int min = 0) { return Math.Min(max, Math.Max(min, val)); }
}

public static class Smarter
{
	public static int ToInt(this string str) { return int.Parse(str); }
	public static long ToLong(this string str) { return long.Parse(str); }
	public static int[] ToInt(this string[] str)
	{
		var r = new int[str.Length];
		for(int i = 0; i < str.Length; i++) r[i] = str[i].ToInt();
		return r;
	 }
}

public class UnionFind
{
	private int[] par;
	public UnionFind(int n)
	{
		par = new int[n];
		for(int i = 0; i < n; i++)
			par[i] = i;
	}
	public int Root(int x)
	{
		if(par[x] == x) return x;
		return par[x] = Root(par[x]);
	}
	public void Unite(int x, int y)
	{
		int rx = Root(x);
		int ry = Root(y);
		if(rx == ry) return;
		par[rx] = ry;
	}
	public bool InSameTree(int x, int y)
	{
		return Root(x) == Root(y);
	}
}