human
Can you provide a Python program to find the number of the cities connected to that city by both roads and railways given the following:
I found an interesting problem on https://atcoder.jp/contests/abc049/tasks/arc065_b:
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-----
4 3 1
1 2
2 3
3 4
2 3

-----Sample Output-----
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.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python program to find the number of the cities connected to that city by both roads and railways.

from collections import Counter

class UnionFind:
  def __init__(self, n):
    self.n = n
    self.par = [i for i in range(n)]
    self.rank = [0]*n
    
    
  def find(self, x):
    if self.par[x] == x:
      return x
    else:
      self.par[x] = self.find(self.par[x])
      return self.par[x]
  
  
  def unite(self, x, y):
    x = self.find(x)
    y = self.find(y)
    if x == y:
      return
    if self.rank[x] < self.rank[y]:
      self.par[x] = y
    else:
      self.par[y] = x
      if self.rank[x] == self.rank[y]:
        self.rank[x] += 1
        
        
  def same(self, x, y):
    return self.find(x) == self.find(y)
  
  
  def members(self, x):
    root = self.find(x)
    return [i for i in range(self.n) if self.find(i) == root]
  

n, k, l  = map(int, input().split())
pq = [list(map(int, input().split())) for i in range(k)]
rs = [list(map(int, input().split())) for i in range(l)]

uf_pq = UnionFind(n)
uf_rs = UnionFind(n)

for p in pq:
  x, y = p
  uf_pq.unite(x-1, y-1)
  
for r in rs:
  x, y = r
  uf_rs.unite(x-1, y-1)
  
pairs = []
for i in range(n):
  pairs.append((uf_pq.find(i), uf_rs.find(i)))
  
count = Counter(pairs)

ans = [count[pair] for pair in pairs]
print(*ans)