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 Python (3.4.3) (judge verdict: Accepted)

from collections import defaultdict
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):  #要素xが属するグループの根を返す
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):  #要素xが属するグループと要素yが属するグループを併合
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):  #グループのサイズ
        return -self.parents[self.find(x)]

    def same(self, x, y):   #要素xと要素yが同じグループかどうか
        return self.find(x) == self.find(y)

    def members(self, x):   #要素xが属するグループに属する要素をリストで返す
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):    #すべての根の要素をリストで返す
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):  #グループの数を変えす
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

n, k, l = map(int, input().split())

uf = UnionFind(n)

p, q = [0] * k, [0] * k
for i in range(k):
    pp, qq = map(int, input().split())
    p[i] = pp - 1 
    q[i] = qq - 1
    uf.union(pp - 1, qq - 1)

uuff = UnionFind(n)
r, s = [0] * l, [0] * l
for i in range(l):
    rr, ss = map(int, input().split())
    r[i] = rr - 1
    s[i] = ss - 1
    uuff.union(rr - 1, ss - 1)
dic = defaultdict(int)
for i in range(n):
    dic[(uf.find(i), uuff.find(i))] += 1

ans = []
for i in range(n):
    ans.append(dic[uf.find(i), uuff.find(i)])

for i in ans:
    print(i,end=" ")