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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;

/*
 <url:https://arc065.contest.atcoder.jp/tasks/arc065_b>
 問題文============================================================
 N 個の都市があり、K 本の道路と L 本の鉄道が都市の間に伸びています。
 i 番目の道路は pi 番目と qi 番目の都市を双方向に結び、
 i 番目の鉄道は ri 番目と si 番目の都市を双方向に結びます。
 異なる道路が同じ 2 つの都市を結ぶことはありません。同様に、異なる鉄道が同じ 2 つの都市を結ぶことはありません。
 
 ある都市から別の都市に何本かの道路を通って到達できるとき、
 それらの都市は道路で連結しているとします。
 また、すべての都市はそれ自身と道路で連結しているとみなします。
 鉄道についても同様に定めます。
 
 全ての都市について、その都市と道路・鉄道のどちらでも連結している都市の数を求めてください。
 
 =================================================================
 
 解説=============================================================
 
 UnionFindを用いて、道路と鉄道、それぞれの集合を作っておく、
 あとは、その集合の根の値を用いて、ある頂点iに関して
 （道路の集合の根、鉄道の集合の根）　が等しいペアを数え上げれば良い
 ================================================================
 */

struct UnionFind {
    vector<int> data;
    UnionFind(int size) : data(size, -1) { }
    bool unionSet(int x, int y) {
        x = root(x); y = root(y);
        if (x != y) {
            if (data[y] < data[x]) swap(x, y);
            data[x] += data[y]; data[y] = x;
        }
        return x != y;
    }
    bool findSet(int x, int y) {
        return root(x) == root(y);
    }
    int root(int x) {
        return data[x] < 0 ? x : data[x] = root(data[x]);
    }
    int size(int x) {
        return -data[root(x)];
    }
};

int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);
    ll N,K,L; cin >> N >> K >> L;
    vector<ll> p(K),q(K),r(L),s(L);
    
    for(int i = 0; i < K;i++){
        cin >> p[i] >> q[i];
        p[i]--; q[i]--;
    }
    for(int i = 0; i < L;i++){
        cin >> r[i] >> s[i];
        r[i]--; s[i]--;
    }
    
    UnionFind pq(N),rs(N);
    for(int i = 0; i < K;i++) pq.unionSet(p[i], q[i]);
    for(int i = 0; i < L;i++) rs.unionSet(r[i], s[i]);
    map<pii,int> mp;
    for(int i = 0; i < N;i++) mp[pii(pq.root(i),rs.root(i))]++;
    for(int i = 0; i < N;i++){
        cout << mp[pii(pq.root(i),rs.root(i))] << " ";
    }
    cout << endl;
    return 0;
}