// Slower C++ solution:
#include <iostream>

#include <bits/stdc++.h>



using namespace std;



int fa[200005];

int fa1[200005];

int vis[200005];



int find1(int a)

{

    if(fa[a]!=a)

        fa[a]=find1(fa[a]);

    return fa[a];

}

void union1(int a,int b)

{

    fa[find1(a)]=find1(b);

}



int find2(int a)

{

    if(fa1[a]!=a)

        fa1[a]=find2(fa1[a]);

    return fa1[a];

}

void union2(int a,int b)

{

    fa1[find2(a)]=find2(b);

}

map<long long,int> mp;

int main()

{

    int n,k,l;

    cin>>n>>k>>l;

    memset(vis,0,sizeof(vis));

    for(int i=0;i<=n;i++)

    {

        fa[i]=i;

        fa1[i]=i;

    }

    for(int i=0;i<k;i++)

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        union1(l1,r);

    }

    for(int i=0;i<l;i++)

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        //if(find1(l1)==find1(r))

        union2(l1,r);

        //cout<<find1(l1)<<' '<<find1(r);

    }

    for(int i=1;i<=n;i++)

    {

        if(mp.find((long long)((long long)find1(i)*200005+(long long)find2(i)))==mp.end())

            mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]=0;

        mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]++;

    }

    for(int i=1;i<n;i++)

        printf("%d ",mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]);

    printf("%d\n",mp[(long long)((long long)find1(n)*200005+(long long)find2(n))]);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7756495360456817x):
#include<ctime>

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstdlib>

#include<cmath> 

#include<cstring> 

#include<cassert>

#include<string>

#include<sstream>

#include<fstream>

#include<deque>

#include<queue>

#include<vector>

#include<map>

#include<list>

#include<stack>

#include<set>

#include<bitset>

#include<iomanip>

#include<utility>

#include<functional>

#include<cctype>

#include<cerrno>

#include<cfloat>

#include<ciso646>

#include<climits>

#include<clocale>

#include<complex>

#include<csetjmp>

#include<csignal>

#include<cstdarg>

#include<cstddef>

#include<cwchar>

#include<cwctype>

#include<exception>

#include<locale>

#include<numeric>

#include<new>

#include<stdexcept>

#include<limits>

using namespace std;



#define ll long long

#define INF 1e9

#define rep(i,n) for(int (i)=0;(i)<n;i++)

#define REP(i,n) for(int (i)=1;(i)<=n;i++)

#define mk(a,b) make_pair(a,b)

#define fi first

#define se second

#define pii pair<int,int>

#define sz(s) s.size()

#define all(s) (s.begin(),s.end())



const int maxn=200005;

int n,k,l;

int f1[maxn],f2[maxn],ans[maxn];

map<pii,vector<int> >mp;





int Find(int f[],int x){

	return f[x]==x?x:f[x]=Find(f,f[x]);

}



void Union(int f[],int x,int y){

	x=Find(f,x);y=Find(f,y);

	if(x!=y)f[x]=y;

}



int main(){

	ios::sync_with_stdio(false);

	cin>>n>>k>>l;

	REP(i,n)f1[i]=f2[i]=i;

	while(k--){

		int u,v;

		cin>>u>>v;

		Union(f1,u,v);

	}

	while(l--){

		int u,v;

		cin>>u>>v;

		Union(f2,u,v);

	}

	REP(i,n)mp[mk(Find(f1,i),Find(f2,i))].push_back(i);

	for(map<pii,vector<int> >::iterator it=mp.begin();it!=mp.end();it++){

		rep(j,it->se.size())ans[it->se[j]]=it->se.size();

	}

	REP(i,n)cout<<ans[i]<<" ";

	return 0;

}