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

#include<queue>

#include<climits>

#include<algorithm>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<string>

#include<ctype.h>

#include<set>

#include<vector>

#include<map>

#include<time.h>

#include<list>

#include<stack>

#include<iostream>

using namespace std;

#define mod 1000000007

#define mem(x,k ) memset(x, (k) ,sizeof(x))

#define mp(A, B) make_pair(A, B)

#define pb push_back

#define sca scanf

#define pri printf

#define fi first

#define se second



typedef long long LL;

typedef long double LD;

typedef pair< int , int  > pii;



const LL INFLL=0x3f3f3f3f3f3f3f3fll;

const int INF=0x7fffffff;

const double PI=acos(-1.0);

const double  eps=1e-9;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

int n,m,k;

int f[200005];

int f2[200005];

map<pair<int ,int >,int> G;

int Father(int k)

{

    if (k==f[k]) return k;

	return f[k]=Father(f[k]);

}



void Union(int x,int y)

{

    if (Father(x)==Father(y)) return;

    else

    {

        f[ f[x] ]=f[y];

        return;

    }

}

int Father2(int k)

{

    if (k==f2[k]) return k;

	return f2[k]=Father2(f2[k]);

}



void Union2(int x,int y)

{

    if (Father2(x)==Father2(y)) return;

    else

    {

        f2[ f2[x] ]=f2[y];

        return;

    }

}

int main(){

    scanf("%d%d%d",&n,&k,&m);

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

        f[i]=f2[i]=i;

    }

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

        int u,v;

        scanf("%d%d",&u,&v);

        Union2(u,v);

    }

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

        int u,v;

        scanf("%d%d",&u,&v);

        Union(u,v);

    }

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

        G[make_pair(Father(i),Father2(i))]++;

    }

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

        //printf("@%d@",Father(i));

        printf("%d",G[make_pair(Father(i),Father2(i))]);

        if (i!=n) printf(" ");

    }

    printf("\n");

    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: