// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <map>

#include <vector>

#include <algorithm>

#include <cstdlib>

#include <cmath>

#include <bitset>

#define pb push_back

#define mp make_pair

#define x first

#define y second

#define pii pair<int,int>

#define pdd pair<db,db>



using namespace std;



typedef long long ll;

typedef double db;



const int N = 1e5+10,M=1e4+10,MOD = 1e9+7;



int fa[N*2],fb[2*N];

//int ans[N*2];



void init(int n){

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

        fa[i]=fb[i]=i;

}



int finda(int x){

    if(fa[x]!=x)fa[x]=finda(fa[x]);

    return fa[x];

}

int findb(int x){

    if(fb[x]!=x)fb[x]=findb(fb[x]);

    return fb[x];

}



int main()

{

    int n,k,l;

    while(~scanf("%d%d%d",&n,&k,&l)){

        init(n);

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

            int x,y;

            scanf("%d%d",&x,&y);

            int fx=finda(x),fy=finda(y);

            if(fx!=fy)fa[fx]=fy;

        }

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

            int x,y;

            scanf("%d%d",&x,&y);

            int fx=findb(x),fy=findb(y);

            if(fx!=fy)fb[fx]=fy;

        }

        map<int,pii> m1;

        map<pii,int> m2;

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

            int xa=finda(i),xb=findb(i);

           //cout<<xa<<" "<<xb<<endl;

            m1[i]=mp(xa,xb);

            m2[mp(xa,xb)]++;

           // cout<<"("<<m1[i].x<<","<<m1[i].y<<") "<<m2[mp(xa,xb)]<<endl<<endl;

        }



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

            cout<<m2[m1[i]]<<" ";

        }

        cout<<endl;

       

    }

}


// 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: