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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
// *テンプレート*
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define P pair<int, int>
// aよりbの方が小さかったら、aの値をbの値に置き換える。(swapではない！)
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
// aよりbの方が大きかったら、aの値をbの値に置き換える。
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

using namespace std;
typedef long long ll;
using Graph = vector<vector<int>>;
//最大公約数
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}
//最小公倍数
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
// 素数かどうか判定
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}

const long long MOD = 1000000007;
const long long INF = 1LL << 60;
// *テンプレートここまで*

// pair -> vector<pair<int,int>> 
//		   AandB.push_back(make_pair(a[i],b[i]))

class Union_Find {
private:
	vector<int>par;
	vector<int>rank;
public:
	Union_Find(int n) {
		rep(i, n) {
			par.push_back(i);
			rank.push_back(0);
		}
	}
	int find(int x) {
		if (par.at(x) == x)return x;
		else return par.at(x) = find(par.at(x));
	}
	void unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y)return;
		if (rank.at(x) < rank.at(y)) {
			par.at(x) = y;
		}
		else {
			par.at(y) = x;
			if (rank.at(x) == rank.at(y))rank.at(x)++;
		}
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
};

signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k, l; cin >> n >> k >> l;
	Union_Find road(n), train(n);
	rep(i, k) {
		int a, b; cin >> a >> b;
		road.unite(a - 1, b - 1);
	}
	rep(i, l) {
		int a, b; cin >> a >> b;
		train.unite(a - 1, b - 1);
	}
	map<P, int>m;
	rep(i, n) {
		m[P(road.find(i), train.find(i))]++;
	}
	rep(i, n - 1) {
		cout << m[P(road.find(i), train.find(i))] << " ";
	}
	cout << m[P(road.find(n - 1), train.find(n - 1))] << endl;
	return 0;
}