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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

//  ____            _                  _                       _       _
// |  _ \ _   _ ___| |_    ___ _ __   | |_ ___ _ __ ___  _ __ | | __ _| |_ ___
// | |_) | | | / __| __|  / __| '_ \  | __/ _ \ '_ ` _ \| '_ \| |/ _` | __/ _ \
// |  _ <| |_| \__ \ |_  | (__| |_) | | ||  __/ | | | | | |_) | | (_| | ||  __/
// |_| \_\\__,_|___/\__|  \___| .__/___\__\___|_| |_| |_| .__/|_|\__,_|\__\___|
//                            |_| |_____|               |_|

//                                                     _        _ _ _____  ___
//                               _ __ ___   __ _ _ __ | |_ __ _/ / |___ / / _ \
//                              | '_ ` _ \ / _` | '_ \| __/ _` | | | |_ \| | | |
//                              | | | | | | (_| | | | | || (_| | | |___) | |_| |
//                              |_| |_| |_|\__,_|_| |_|\__\__,_|_|_|____/ \___/

//https://github.com/manta1130/Competitive_Programming_Template_Rust

#[macro_use]
mod input {

    use std;
    use std::io;

    const SPLIT_DELIMITER: char = ' ';

    #[macro_export]
    #[allow(unused_macros)]
    macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
}

    #[allow(dead_code)]
    pub fn input_line_str() -> String {
        let mut s = String::new();
        io::stdin().read_line(&mut s).unwrap();
        s.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn p<T>(t: T)
    where
        T: std::fmt::Display,
    {
        println!("{}", t);
    }

    #[allow(dead_code)]
    pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<Vec<T>> = Vec::new();

        for _ in 0..line {
            let vec_line = input_vector();
            v.push(vec_line);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector<T>() -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<T> = Vec::new();

        let s = input_line_str();
        let split_result = s.split(SPLIT_DELIMITER);
        for z in split_result {
            let buf = match z.parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector_row<T>(n: usize) -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v = Vec::with_capacity(n);
        for _ in 0..n {
            let buf = match input_line_str().parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    pub trait ToCharVec {
        fn to_charvec(&self) -> Vec<char>;
    }

    impl ToCharVec for String {
        fn to_charvec(&self) -> Vec<char> {
            self.to_string().chars().collect::<Vec<_>>()
        }
    }
}

mod union_find {

///素集合データ構造
pub struct UnionFind {
    v: Vec<usize>,
}

impl UnionFind {
    ///要素数nの素集合データ構造を構築する。
    pub fn new(n: usize) -> UnionFind {
        let mut v = Vec::new();
        for i in 0..n {
            v.push(i);
        }
        UnionFind { v: v }
    }

    ///親を探索する。
    pub fn find(&mut self, a: usize) -> usize {
        if a == self.v[a] {
            return a;
        }

        let buf = self.v[a];
        let parent = self.find(buf);
        self.v[a] = parent;
        parent
    }

    ///親が同じか判定する。
    pub fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    ///結合する。
    pub fn union(&mut self, a: usize, b: usize) {
        let parent_a = self.find(a);
        let parent_b = self.find(b);
        self.v[parent_a] = parent_b;
    }
}
}

use input::*;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use union_find::*;

fn main() {
    let (n, k, l): (usize, usize, usize);
    input!(n, k, l);
    let mut uft = UnionFind::new(n);
    let mut ufr = UnionFind::new(n);
    let mut setmapr = BTreeMap::new();
    let mut setmapt = BTreeMap::new();

    for _ in 0..k {
        let (a, b): (usize, usize);
        input!(a, b);
        uft.union(a - 1, b - 1);
    }
    for _ in 0..l {
        let (a, b): (usize, usize);
        input!(a, b);
        ufr.union(a - 1, b - 1);
    }

    for i in 0..n {
        let parent = uft.find(i);
        let po = setmapt.entry(parent).or_insert(BTreeSet::new());
        po.insert(i);
    }
    for i in 0..n {
        let parent = ufr.find(i);
        let po = setmapr.entry(parent).or_insert(BTreeSet::new());
        po.insert(i);
    }
    for i in 0..n {
        let parentt = uft.find(i);
        let parentr = ufr.find(i);
        print!("{} ", (&setmapr[&parentr] & &setmapt[&parentt]).len());
    }
    p("");
}