// Slower C++ solution:
//#include <bits/stdc++.h>

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <math.h>

#include <stdlib.h>

#include <limits.h>

#include <algorithm>

#include <queue>

#include <vector>

#include <set>

#include <map>

#include <stack>

#include <bitset>

#include <string>

#include <time.h>

using namespace std;

long double esp=1e-11;

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#define fi first

#define se second

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

#define cle(a) while(!a.empty())a.pop()

#define mem(p,c) memset(p,c,sizeof(p))

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

#define pb push_back

#define lson l , m , rt << 1

#define rson m + 1 , r , rt << 1 | 1

typedef long long int LL;

const long double PI = acos((long double)-1);

const LL INF=0x3f3f3f3f3f3f3f3fll;

const int MOD =1000000007ll;

const int maxn=210100;

int ro1[maxn];

int up1(int u){return(ro1[u]==u)?u:ro1[u]=up1(ro1[u]);}

int ro2[maxn];

int up2(int u){return(ro2[u]==u)?u:ro2[u]=up2(ro2[u]);}

vector<pair<int,int> >node[maxn];

int c[maxn][2];

int ans[maxn];

int main()

{

    //freopen("in9.txt", "r", stdin);

    //freopen("in9.txt", "r", stdin);

    //freopen("out9.txt", "w", stdout);

    //::iterator iter;                  %I64d

    //for(int x=1;x<=n;x++)

    //for(int y=1;y<=n;y++)

    //scanf("%d",&a);

    //printf("%d\n",ans);

    int n,k,l;

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

    for(int x=1;x<=n;x++)ro1[x]=ro2[x]=x,node[x].clear();

    for(int x=1;x<=k;x++)

    {

        int u,v;

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

        ro1[up1(u)]=up1(v);

    }

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

        node[up1(x)].pb(mp(x,0));

    for(int x=1;x<=l;x++)

    {

        int u,v;

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

        ro2[up2(u)]=up2(v);

    }

    mem(ans,0);

    mem(c,0);

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

    {

    	for(int y=0;y<node[x].size();y++)

		{

			node[x][y].se=up2(node[x][y].fi);

			if(c[node[x][y].se][1]!=x)

			{

				c[node[x][y].se][1]=x;

				c[node[x][y].se][0]=0;

			}

			c[node[x][y].se][0]++;

		}

		for(int y=0;y<node[x].size();y++)

			ans[node[x][y].fi]=c[node[x][y].se][0];

    }

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

        printf("%d%c",ans[x]," \n"[x==n]);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 49.71270349037732x):
#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]);

}