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: Wrong Answer)

using System;
using System.Collections;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Collections.Generic;
using System.Threading.Tasks;

class Program
{
	static int N, K, L;
	//所謂Union Find木を配列で表現
	//それぞれインデックスの根を格納する
	static int[] ufRoad;
	static int[] ufRail;

	//Main
	static void Main(string[] args)
	{
		//読み込む
		string[] S = Console.ReadLine().Split(' ');
		N = int.Parse(S[0]);
		K = int.Parse(S[1]);
		L = int.Parse(S[2]);

		//union find木
		//便宜上、インデックス0は使用しない
		ufRoad = new int[N+1];
		ufRail = new int[N+1];
		//自身を根とする状態（みんなバラバラの状態）で初期化
		for(int i=0; i<N+1; i++){
				ufRoad[i] = i;
				ufRail[i] = i;
		}

		//読み込む
		for(int i=0; i<K; i++){
				string[] line = Console.ReadLine().Split(' ');
				int p = int.Parse(line[0]);
				int q = int.Parse(line[1]);
				//条件より、p<q
				//より小さい方を根とすることとする
				ufRoad[q] = p;
		}
		for(int i=0; i<L; i++){
				string[] line = Console.ReadLine().Split(' ');
				int r = int.Parse(line[0]);
				int s = int.Parse(line[1]);
				//条件より、r<s
				//より小さい方を根とすることとする
				ufRail[s] = r;
		}

		//上から順に根を確定
		for(int i=1; i<N+1; i++){
				if(i == ufRoad[i]){
						continue;
				}
				else{
						int par = ufRoad[i];
						ufRoad[i] = ufRoad[par];
				}
				if(i == ufRail[i]){
						continue;
				}
				else{
						int par = ufRail[i];
						ufRail[i] = ufRail[par];
				}
		}

		/*デバッグ
		for(int i=1; i<N+1; i++){
				Console.Write(i + "," + ufRoad[i] + " ");
		}
		Console.WriteLine(" ");
		for(int i=1; i<N+1; i++){
				Console.Write(i + "," + ufRail[i] + " ");
		}
		*/

		//特定の根のペアの個数を格納するディクショナリ
		Dictionary<string,int> pairDic = new Dictionary<string,int>();

		for(int i=1; i<N+1; i++){
				string key = ufRoad[i].ToString() + "," + ufRail[i].ToString();
				//既に当該keyが存在する場合
				if(pairDic.ContainsKey(key)){
						//valueを加算
						pairDic[key] ++;
				}
				//まだ存在しない場合
				else{
						pairDic.Add(key, 1);
				}
		}

		//ペアの個数を出力
		//1 ~ N-1まで
		for(int i=1; i<N; i++){
				string key = ufRoad[i].ToString() + "," + ufRail[i].ToString();
				Console.Write(pairDic[key] + " ");
		}
		//Nの場合
		string keyN = ufRoad[N].ToString() + "," + ufRail[N].ToString();
		Console.WriteLine(pairDic[keyN]);
	}
}