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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N,K,L = map(int, input().split())
pq_list = []
for _ in range(K):
    pq_list.append(list(map(int, input().split())))

rs_list = []
for _ in range(L):
    rs_list.append(list(map(int, input().split())))

class UnionFind():
    def __init__(self):
        self.roots = {}
        self.ranks = {}

    def connect(self,x,y):
        root_x = self.pick_root(x)
        root_y = self.pick_root(y)
        if not root_x == root_y:
            if self.ranks[root_x] >= self.ranks[root_y]:
                # 短い方の根を長い方の根にする
                # 等しいときはxにくっつける
                self.roots[y] = root_x
            else:
                self.roots[x] = root_y

        if self.ranks[root_x]==self.ranks[root_y]:
            # 等しいときxにくっつけると、xのランクが1増える
            self.ranks[x] += 1

    def pick_root(self,k):
        if not k in self.roots:
            self.roots[k] = k
            self.ranks[k] = 1
            return k

        k_root = self.roots[k]
        if k_root == k:
            return k
        else:
            # 経路圧縮（のつもり）
            # pick_rootで毎回見に行くのが無駄にも思えるが、いつ根ノードがくっつくかわからないので必要な気がしている
            self.roots[k] = self.pick_root(self.roots[k_root])
            return self.roots[k]
#             return self.pick_root(self.roots[k])

UF_pq = UnionFind()
for p,q in pq_list:
    UF_pq.connect(p, q)

UF_rs = UnionFind()
for r,s in rs_list:
    UF_rs.connect(r, s)

root_tuples = []
for i in range(1,N+1):
    if not i in UF_pq.roots:
        root_tuples.append((None,None))
        continue
    if not i in UF_rs.roots:
        root_tuples.append((None,None))
        continue
    pq_root = UF_pq.roots[i]
    rs_root = UF_rs.roots[i]
    root_tuples.append((pq_root,rs_root))

from collections import Counter
counter = Counter(root_tuples)
counter[(None,None)] = 1

kotae_list = []
for i in range(N):
    kotae_list.append(counter[root_tuples[i]])

print(" ".join(map(str,kotae_list)))