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.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class Main{

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        int L = sc.nextInt();
        DJSet road = new DJSet(N+1);
        DJSet rail = new DJSet(N+1);
        for (int i = 0; i < K; i++) {
            int p = sc.nextInt();
            int q = sc.nextInt();
            road.union(p, q);
        }
        for (int i = 0; i < L; i++) {
            int r = sc.nextInt();
            int s = sc.nextInt();
            rail.union(r, s);
        }
        HashMap<Long, Integer> map = new HashMap<Long, Integer>();
        long[] keys = new long[N+1];
        for (int i = 1; i <= N; i++) {
            long key = (long)road.root(i) * (N+1) + rail.root(i);
            keys[i] = key;
            if(map.containsKey(key)) map.replace(key, map.get(key) + 1);
            else map.put(key, 1);
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 1; i <= N; i++) {
            sb.append(map.get(keys[i]));
            if(i < N) sb.append(" ");
        }

        System.out.println(sb.toString());
        sc.close();
    }

}

class DJSet {
    /**
     * 各要素の親を保持する配列。
     * 自信が根である要素は負の数を持つ。
     * （負の数の場合、絶対値がその集合の要素数）
     */
    public int[] upper;

    /**
     * コンストラクタ。
     * @param n 対象となる要素数
     */
    public DJSet(int n){
        upper = new int[n];
        Arrays.fill(upper, -1);
    }

    /**
     * 要素xが含まれる集合のrootを求める。
     * rootを求める過程で経路圧縮しrootが親になるようにする。
     * @param x 要素x
     * @return 要素xが含まれる集合のroot
     */
    public int root(int x){
        return upper[x] < 0 ? x : (upper[x] = root(upper[x]));
    }

    /**
     * 要素xと要素yが同じ集合か判定する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yが同じ集合であればtrue、そうでなければfalseを返す。
     */
    public boolean same(int x, int y){
        return root(x) == root(y);
    }

    /**
     * 要素xと要素yを含むそれぞれの集合を合併する。
     * 要素数の多い集合のrootを残し、少ないほうのrootをその下に合併する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yを含むそれぞれの集合を合併できたらtrue、もともと同じ集合で合併しなかったらfalseを返す。
     */
    public boolean union(int x, int y){
        x = root(x);
        y = root(y);
        if(x != y){
            if(upper[y] < upper[x]){
                int t = x;
                x = y;
                y = t;
            }
            upper[x] += upper[y];
            upper[y] = x;
        }
        return x != y;
    }

    /**
     * 集合の数を数える。
     * @return 集合の数
     */
    public int countSet(){
        int c = 0;
        for(int u :upper){
            if(u < 0) c++;
        }
        return c;
    }

    /**
     * 要素ｘが含まれる集合の要素数を数える。
     * @param x 要素ｘ
     * @return 要素ｘが含まれる集合の要素数
     */
    public int countElement(int x){
        return root(x) * -1;
    }
}