// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

struct abc {

	int aa, bb, cc;

	bool operator < (const abc&T) const {

		return(aa < T.aa || (aa == T.aa && bb < T.bb) || (aa == T.aa && bb == T.bb && cc < T.cc));

	}

	bool operator == (const abc&T) const {

		return(aa == T.aa && bb == T.bb);

	}

} ans[200005];

using namespace std;

int N, KK, LL, i, j, k, l, m, n, pt1[200005], pt2[200005], s1[200005], s2[200005], z;

int sz[200005], ans2[200005];

int find1(int aa) {

	if(pt1[aa] != aa) 

		pt1[aa] = find1(pt1[aa]);

	return(pt1[aa]);

}

int find2(int bb) {

	if(pt2[bb] != bb) 

		pt2[bb] = find2(pt2[bb]);

	return(pt2[bb]);

}

int main() {

	scanf("%d%d%d", &N, &KK, &LL);

	for(i = 1; i <= N; i++) {

		pt1[i] = i;

		pt2[i] = i;

		s1[i] = 1;

		s2[i] = 1;

	}

	for(i = 0; i < KK; i++) {

		scanf("%d%d", &j, &l);

		if(find1(j) == find1(l)) continue;

		s1[find1(l)] += s1[find1(j)];

		pt1[find1(j)] = find1(l);

	}

	

	for(i = 0; i < LL; i++) {

		scanf("%d%d", &j, &l);

		if(find2(j) == find2(l)) continue;

		s2[find2(l)] += s2[find2(j)];

		pt2[find2(j)] = find2(l);

	}

	for(i = 1; i <= N; i++) {

		pt1[i] = find1(i);

		pt2[i] = find2(i);

		ans[i].aa = pt1[i];

		ans[i].bb = pt2[i];

		ans[i].cc = i;

	}

//	for(i = 1; i <= N; i++) printf("%d %d\n", pt1[i], pt2[i]);

	

	sort(ans + 1, ans + N + 1);

	sz[1] = 1;

	for(i = 2; i <= N; i++) if(ans[i] == ans[i - 1]) {

		sz[i] = sz[i - 1] + 1;

	} else {

		sz[i] = 1;

	}

	for(i = N - 1; i >= 1; i--) 

	if(ans[i] == ans[i + 1]) sz[i] = sz[i + 1];

	

	//for(i = 1; i <= N; i++) printf("%d %d %d %d\n", ans[i].aa, ans[i].bb, ans[i].cc, sz[i]);

	

	for(i = 1; i <= N; i++) ans2[ans[i].cc] = sz[i];

	

	for(i = 1; i < N; i++) printf("%d ", ans2[i]);

	printf("%d\n", ans2[N]);

}

// Fastest known solution for this problem: