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: Time Limit Exceeded)

import sys
stdin = sys.stdin

sys.setrecursionlimit(10**5)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

class UnionFind:
    def __init__(self, node:int) -> None:
        self.n = node
        self.par = [i for i in range(self.n)]
        self.rank = [0 for i in range(self.n)]
        
    def find(self, x:int) -> int:
        if x == self.par[x]:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]
        
    def unite(self, x:int, y:int) -> bool:
        if self.isSame(x,y):
            #print("x and y has already united")
            return False
        
        rx = self.find(x)
        ry = self.find(y)
        
        if self.rank[rx] < self.rank[ry]:
            self.par[rx] = self.par[ry]
        else:
            self.par[ry] = self.par[rx]
            
            if self.rank[rx] == self.rank[ry]:
                self.rank[rx] += 1
                
        return True
                
    def isSame(self, x:int, y:int) -> bool:
        return self.find(x) == self.find(y)
    
n,k,l = li()
pq = []
rs = []

for _ in range(k):
    pq.append(tuple(li_()))
    
for _ in range(l):
    rs.append(tuple(li_()))

road = UnionFind(n)
rail = UnionFind(n)

for p,q in pq:
    road.unite(p,q)
    
for r,s in rs:
    rail.unite(r,s)
    
for i in range(n):
    road.find(i)
    
for i in range(n):
    rail.find(i)
    
road_dic = {i:set() for i in range(n)}
rail_dic = {i:set() for i in range(n)}

for i in range(n):
    road_dic[road.par[i]].add(i)
    rail_dic[rail.par[i]].add(i)
    
ans = []
for i in range(n):
    ans.append(len(road_dic[road.par[i]] & rail_dic[rail.par[i]]))
    
print(*ans)