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

#include <cstring>

#include <algorithm>

#include <vector>

#define ll long long

using namespace std;



const int maxn = 200010;

int n, m1, m2, u, v;

int f1[maxn], f2[maxn], vis[maxn], cnt[maxn], res[maxn];

vector <int> G[maxn];



int Find1 (int x) {

	if (f1[x] == x) return x;

	return f1[x] = Find1(f1[x]);

}



int Find2 (int x) {

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

	return f2[x] = Find2(f2[x]);

}



int main () {

	scanf ("%d%d%d", &n, &m1, &m2);

	for (int i = 1; i <= n; i++) f1[i]= f2[i] = i;

	for (int i = 0; i < m1; i++) {

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

        f1[Find1(u)] = Find1(v);

	}

	for (int i = 0; i < m2; i++) {

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

        f2[Find2(u)] = Find2(v);

	}

    for (int i = 1; i <= n; i++) G[Find1(i)].push_back(i);

    for (int i = 1; i <= n; i++) if(G[i].size()) {

        for (int j = 0; j < G[i].size(); j++) {

            int t = Find2(G[i][j]);

            if (vis[t] != i) {

                vis[t] = i;

                cnt[t] = 0;

            }

            cnt[t]++;

        }

        for (int j = 0; j < G[i].size(); j++) {

            int t = Find2(G[i][j]);

            res[G[i][j]] = cnt[t];

        }

    }

    for (int i = 1; i <= n; i++) printf("%d%c", res[i], i==n?'\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: