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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
 
 
public class Main {
 
	static long MOD = 1_000_000_007;
	static int INF = 1_000_000_007;
	//二項係数関連ここから-----
	static boolean isPreprocessing = false;	//二項係数を計算するための前処理を行っているかどうかのフラグ
	static int MAX = 260;				//MAXまで階乗やそれらの逆元のテーブルを保持する。
	static long[] fac;	//fac[i] := i! % MOD
	static long[] inv;	//inv[i] := mod. MODにおける i の逆元
	static long[] finv;	//finv[i] := mod. MODにおける i! の逆元(invの累積"積")
	//二項係数関連おわり------
	
	static int[] roots;		//Union-find木用
	static int[] ans;
	static List<ArrayList<Integer>> G;
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		//int N = Integer.parseInt(sc.next());
		//long L = Long.parseLong(sc.next());
		//char[] A = sc.next().toCharArray();
		int N = Integer.parseInt(sc.next());
		int K = Integer.parseInt(sc.next());
		int L = Integer.parseInt(sc.next());
		UnionFind uf_road = new UnionFind(N);
		UnionFind uf_train = new UnionFind(N);
		
		for(int i=0; i<K; i++){
			int p = Integer.parseInt(sc.next())-1;
			int q = Integer.parseInt(sc.next())-1;
			uf_road.join(p, q);
		}
		for(int i=0; i<L; i++){
			int r = Integer.parseInt(sc.next())-1;
			int s = Integer.parseInt(sc.next())-1;
			uf_train.join(r, s);
		}
		for(int i=0; i<N; i++){
			uf_road.root(i);
			uf_train.root(i);
		}
		class City{
			private int root_road;
			private int root_train;
			private int number;
			public City(int root_road, int root_train, int number){
				this.number = number;
				this.root_road = root_road;
				this.root_train = root_train;
			}
			public int getRoot_road() {
				return root_road;
			}
			public int getNumber() {
				return number;
			}
			public int getRoot_train() {
				return root_train;
			}

		}
		City[] data = new City[N];
		for(int i=0; i<N; i++){
			data[i] = new City(uf_road.getRoots()[i], uf_train.getRoots()[i], i);
		}
		Comparator<City> c = Comparator.comparing(City::getRoot_road).thenComparing(City::getRoot_train);
		Arrays.sort(data, c);
		
		City tmp = data[0];
		int count = 0;
		int[] ans = new int[N];
		int first = 0;
		for(int i=0; i<N; i++){
			if(data[i].getRoot_road() == tmp.getRoot_road() && data[i].getRoot_train() == tmp.getRoot_train()){
				count++;
			}else{
				for(int j=first; j<first+count; j++){
					ans[data[j].getNumber()] = count;
				}
				count = 1;
				tmp = data[i];
				first = i;
			}
		}
		for(int j=first; j<first+count; j++){
			ans[data[j].getNumber()] = count;
		}
		for(int a:ans){
			System.out.print(a+" ");
		}
		
	}
	
	//Union-Find木
	static class UnionFind{
		private int[] roots;
		public UnionFind(int N){
			this.roots = new int[N];
			for(int i=0; i<N; i++){
				this.roots[i] = i;
			}
		}
		
		public boolean same(int a, int b){
			if(root(a) == root(b)){
				return true;
			}else{
				return false;
			}
		}
		
		public int root(int a){
			if(a == this.roots[a])return a;
			return this.roots[a] = root(this.roots[a]);
		}
		
		public void join(int a, int b){
			if(root(a) == root(b))return;
			this.roots[root(b)] = root(a);
		}
		
		public int[] getRoots(){
			return this.roots;
		}
	}

	//素因数分解をする(index_0:素数 index_1:個数)
	static List<long[]> factorization(long M){
		List<long[]> primes = new ArrayList<>();
 
		long now_prime = 2;
		long now_M = M;
		while(now_prime*now_prime<=M){
			if(now_M%now_prime == 0){
				now_M /= now_prime;
				long[] prime = new long[2];
				prime[0] = now_prime;
				prime[1] = 1;
				while(now_M%now_prime == 0){
					now_M /= now_prime;
					prime[1]++;
				}
				primes.add(prime);
				
			}
			if(now_prime%2 == 0){
				now_prime++;
			}else{
				now_prime += 2;
			}
		}
		if (now_M>1 || M==1){
			long[] prime = new long[2];
			prime[0] = now_M;
			prime[1] = 1;
			primes.add(prime);
		}
		
		return primes;
	}
	
	//２つのint配列を連結させる
	static int[] joinArray(int[] A, int[] B){
		int[] tmp = new int[A.length + B.length];
		System.arraycopy(A, 0, tmp, 0, A.length);
		System.arraycopy(B, 0, tmp, A.length, B.length);
		return tmp;
	}
	
	//２つのint配列が等しいかどうか判定する
	static boolean arrayEqual(int[] A, int[] B){
		if(A.length != B.length)return false;
		for(int i=0; i<A.length; i++){
			if(A[i] != B[i])return false;
		}
		return true;
	}
	
	//int配列をindex順に見て、要素の重複があった場合その要素と間の要素を削除する12123→23
	static int[] calc(int[] A){
		boolean[] tmp = new boolean[A.length];
		Arrays.fill(tmp, true);
		int fetch = 0;
		while(fetch<A.length-1){
			boolean isExist = false;
			for(int i=fetch+1; i<A.length;i++){
				if(A[fetch] == A[i]){
					for(int j=fetch;j<=i;j++){
						tmp[j]=false;
					}
					fetch = i+1;
					isExist = true;
					break;
				}
			}
			if(!isExist)fetch++;
		}
		List<Integer> list = new ArrayList<>();
		for(int i=0; i<A.length; i++){
			if(tmp[i])list.add(A[i]);
		}
		
		int[] foo = new int[list.size()];
		for(int i=0; i<list.size(); i++){
			foo[i] = list.get(i);
		}
		/*
		for(boolean t : tmp)
			System.out.print(t+" ");
		*/
		return foo;
	}
	
	
	//aとbの最小公倍数を返す
	static long lcm (long a, long b) {
		long temp;
		long c = a;
		c *= b;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return (c/b);
	}
	
	//二項係数用の前処理
	static void COMinit(){
		//二項係数演算用前処理
		fac = new long[MAX+1];
		fac[0] = 1L;
		fac[1] = 1L;
		inv = new long[MAX+1];
		inv[1] = 1L;
		finv = new long[MAX+1];
		finv[0] = 1L;
		finv[1] = 1L;
		for(int i=2;i<MAX;i++){
			fac[i] = fac[i-1] * i % MOD;
			inv[i] = MOD - inv[(int)(MOD%i)] * (MOD/i) % MOD;
			finv[i] = finv[i-1] * inv[i] % MOD;
		}
		isPreprocessing = true;
	}
	
	//二項係数i_C_jをMODで割った余りを返す
	static long COM(int i, int j){
		if(i<j)return 0;
		if(i<0 || j<0)return 0;
		if(!isPreprocessing)COMinit();
		return fac[i] * (finv[j] * finv[i-j] % MOD) % MOD;
	}
	
}