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 (1.42.0) (judge verdict: Time Limit Exceeded)

#![allow(unused_imports)]
#![allow(clippy::many_single_char_names)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repo' is at "https://github.com/Loptall/prcn_lib"

/*
    そう甘くはないんだな〜w
*/

// from std...
use std::cmp::{
    max, min, Ordering,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::convert::TryInto;
use std::mem::swap;
use std::num::{NonZeroU32, ParseIntError};
use std::ops::{
    Add, AddAssign, BitAnd, BitAndAssign, BitOr, BitOrAssign, BitXor, BitXorAssign, Div, DivAssign,
    Index, IndexMut, Mul, MulAssign, Neg, Rem, RemAssign, Shl, ShlAssign, Shr, ShrAssign, Sub,
    SubAssign,
};
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

// Dep' crates are...
use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, multinomial, Integer};
use num_rational::Rational;
use num_traits::{clamp, one, pow, zero, Num, NumAssignOps, NumOps, One, Pow, Zero};
use permutohedron::Heap;
use proconio::{
    fastout, input, is_stdin_empty,
    marker::{Bytes, Chars, Isize1, Usize1},
};
use rand::random;

pub const MOD10E9_7: usize = 1000000007;
pub const MOD99_: usize = 998244353;
pub const MAX: usize = std::usize::MAX;
pub const PI: f64 = f64::consts::PI;
pub const A: u8 = 65;

#[derive(Debug, Clone)]
pub struct UnionFind {
    parent: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    pub fn new(n: usize) -> Self {
        Self {
            parent: (0..n).collect::<Vec<usize>>(),
            size: vec![1; n],
        }
    }

    pub fn len(&self) -> usize {
        self.parent.len()
    }

    /// i が属する集合の親のインデックス
    pub fn find(&mut self, mut i: usize) -> usize {
        while self.parent[i] != i {
            self.parent[i] = self.find(self.parent[i]);
            i = self.parent[self.parent[i]];
        }
        i
    }

    /// a と b を繋ぐ
    pub fn unite(&mut self, a: usize, b: usize) -> bool {
        let mut a = self.find(a);
        let mut b = self.find(b);
        if a == b {
            return false;
        }
        if self.size[a] < self.size[b] {
            std::mem::swap(&mut a, &mut b);
        }
        self.size[a] += self.size[b];
        self.parent[b] = a;
        true
    }

    /// i が属する集合の要素数
    pub fn count(&mut self, i: usize) -> usize {
        let p = self.find(i);
        self.size[p]
    }


    /// a と b が同一集合に属するか
    pub fn joint(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    /// i が属する集合の要素を列挙する
    pub fn get_group(&mut self, i: usize) -> HashSet<usize> {
        let p = self.find(i);
        (0..self.len()).filter(|x| self.find(*x) == p).collect()
    }
}

// code...
// #[fastout]
fn main() {
    input! {
        n: usize, k: usize, l: usize,
        pq: [(Usize1, Usize1); k],
        rs: [(Usize1, Usize1); l]
    }

    let mut road = UnionFind::new(n);
    let mut train = UnionFind::new(n);

    for (p, q) in pq {
        road.unite(p, q);
    }

    for (r, s) in rs {
        train.unite(r, s);
    }

    for i in 0..n {
        let ans = road.get_group(i).intersection(&train.get_group(i)).count();
        print!("{} ", ans);
    }
}