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

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <assert.h>

#define IOS ios::sync_with_stdio(false)

using namespace std;

#define inf (0x3f3f3f3f)

typedef long long int LL;





#include <iostream>

#include <sstream>

#include <vector>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <bitset>

const int maxn = 2e5 + 20;

int fa[2][maxn];

int getFa(int u, int which) {

    if (fa[which][u] == u) return u;

    else return fa[which][u] = getFa(fa[which][u], which);

}

void toMerge(int x, int y, int which) {

    x = getFa(x, which);

    y = getFa(y, which);

    fa[which][y] = x;

}

map<pair<int, int>, int>al;

void work() {

    int n, k1, k2;

    scanf("%d%d%d", &n, &k1, &k2);

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

        fa[0][i] = fa[1][i] = i;

    }

    for (int i = 1; i <= k1; ++i) {

        int u, v;

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

        toMerge(u, v, 0);

    }

    for (int i = 1; i <= k2; ++i) {

        int u, v;

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

        toMerge(u, v, 1);

    }

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

        al[make_pair(getFa(i, 0), getFa(i, 1))]++;

    }

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

        cout << al[make_pair(getFa(i, 0), getFa(i, 1))] << " ";

    }

}



int main() {

#ifdef local

    freopen("data.txt", "r", stdin);

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

#endif

    work();

    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: