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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<utility>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<queue>
#include<set>
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define prin(arg) std::cout<<arg<<"\n"
#define prin2(arg1,arg2) std::cout<<arg1<<" "<<arg2<<"\n"
#define fill(arg,n) std::memset(arg,n,sizeof(arg))
#define mp std::make_pair
#define pb push_back
using std::cin;
typedef long long ll;
typedef std::pair<int,int> pi;
typedef std::vector<int> vi;
typedef std::set<int> si;
typedef std::string str;
const int INF=1e+9;
const ll INFLL=1e+17;
const ll MOD=1e+9+7;
const int NMAX=200010;
int N,K,L;
int p[100010],q[100010],r[100010],s[100010];
//UnionFindTreeのクラス。
/***************************/
class uft{
public:
  int par[NMAX];//親
  int rank[NMAX];//木の深さ
  int size[NMAX];//木の大きさ
  uft(int n);//コンストラクタ
  int find(int x);
  void unite(int x,int y);
  bool same(int x,int y);
  int usize(int x);
};
uft::uft(int n){//コンストラクタ uft uf(N)でサイズNのunionfindtreeを宣言 
  for(int i=0;i<n;i++){
    par[i]=i;
    rank[i]=0;
    size[i]=1;
  }
}
int uft::find(int x){//xの根を返す
  if(par[x]==x) return x;//根の親は自分自身
  else return par[x]=find(par[x]);//再起で親を辿る
}
void uft::unite(int x,int y){//x,yを併合
  x=find(x);
  y=find(y);
  if(x==y) return ;//根が同じなら併合する必要なし

  if(rank[x]<rank[y]){//木の深い方に浅い方をつける
    par[x]=y;
    size[y]+=size[x];
  }
  else{
    par[y]=x;
    size[x]+=size[y];
    if(rank[x]==rank[y]) rank[x]++;//同じ深さなら深さを増やす
  }
}
bool uft::same(int x,int y){//x,yが同じグループにいるかどうかを返す
  return find(x)==find(y);//根が同じかどうかを判断する
}
int uft::usize(int x){//xの属する木のサイズを返す
  return size[find(x)];
}

int main(){
  cin>>N>>K>>L;
  rep(i,K){
    cin>>p[i]>>q[i];
    p[i]--;q[i]--;
  }
  rep(i,L){
    cin>>r[i]>>s[i];
    r[i]--;s[i]--;
  }
  uft ur(N),ut(N);
  rep(i,K){
    ur.unite(p[i],q[i]);
  }
  rep(i,L){
    if(ur.same(r[i],s[i])){
      ut.unite(r[i],s[i]);
    }
  }
  rep(i,N){
    std::cout<<ut.usize(i)<<" ";
  }
  std::cout<<"\n";
  return 0;
}