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

#include <cstdio>

#include <vector>

#include <map>

#include <algorithm>

#include <cstring>

using namespace std;

typedef long long LL;

const int maxn = 4e5 + 8;



//有时可能需要 并查集+离散化





int father[maxn], _rank[maxn];

inline void DisjointSet(int n)

{

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

        father[i] = i;

        _rank[i] = 0;

    }

}



inline int _find(int v)

{

    return father[v] = father[v] == v ? v : _find(father[v]);

    //同时压缩路径，有时不能压缩路径，有时必须压缩路径，看具体情况

}



inline void _merge(int x, int y)

{

    int a = _find(x), b = _find(y);                //

    if(_rank[a] < _rank[b]){

        father[a] = b;

    }

    else{

        father[b] = a;

        if(_rank[a] == _rank[b]){

            _rank[a]++;

        }

    }

}





int father1[maxn], _rank1[maxn];

inline void DisjointSet1(int n)

{

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

        father1[i] = i;

        _rank1[i] = 0;

    }

}



inline int _find1(int v)

{

    return father1[v] = father1[v] == v ? v : _find1(father1[v]);

    //同时压缩路径，有时不能压缩路径，有时必须压缩路径，看具体情况

}



inline void _merge1(int x, int y)

{

    int a = _find1(x), b = _find1(y);                //

    if(_rank1[a] < _rank1[b]){

        father1[a] = b;

    }

    else{

        father1[b] = a;

        if(_rank1[a] == _rank1[b]){

            _rank1[a]++;

        }

    }

}





map<int, vector<int>> mp1, mp2;

map<int, int> cnt;

int ans[maxn];



int main()

{

    #ifdef LOCAL

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

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

    int T = 4;

    while(T--){

    #endif // LOCAL

    ios::sync_with_stdio(false); cin.tie(0);





    int n, p, q, u, v;

    cin >> n >> p >> q;

    DisjointSet(n), DisjointSet1(n);

    while(p--){

        cin >> u >> v;

        if(_find(u) != _find(v)) _merge(_find(u), _find(v));

    }



    while(q--){

        cin >> u >> v;

        if(_find1(u) != _find1(v)) _merge1(_find1(u), _find1(v));

    }



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

        mp1[_find(i)].push_back(i);

    }

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

        mp2[_find1(i)].push_back(i);

    }



    for(auto i = mp1.begin(); i != mp1.end(); i++){

        sort((i->second).begin(), (i->second).end());

    }

    for(auto i = mp2.begin(); i != mp2.end(); i++){

        sort((i->second).begin(), (i->second).end());

    }



    //cout << "?"<<endl;



    int i, j, a, b, pa, pb, sza, szb, l, r, mid;

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

        if(ans[i] != 0) continue;

        pa = _find(i); pb = _find1(i);

        sza = mp1[pa].size(), szb = mp2[pb].size();

        cnt.clear();

        a = 0, b = 0;

        while(a < sza && b < szb){

            if(mp1[pa][a] == mp2[pb][b]){

                cnt[mp1[_find(i)][a]]++;

                a++;

                b++;

            }

            else if(mp1[pa][a] > mp2[pb][b]){

                if(b + 1== szb){

                    break;

                }

                else if(mp1[pa][a] > mp2[pb][szb - 1]){

                    break;

                }

                else{

                    l = b, r = szb;

                    b = szb - 1;

                    while(l + 1 < r){

                        mid = (l + r) >> 1;

                        if(mp1[pa][a] <= mp2[pb][mid]){

                            r = mid;

                            b = min(b, mid);

                        }

                        else l = mid;

                    }



                }

            }

            else{

                if(a + 1 == sza){

                    break;

                }

                else if(mp1[pa][sza - 1] < mp2[pb][b]){

                    break;

                }

                else{

                    l = a, r = sza;

                    a = sza - 1;

                    while(l + 1 < r){

                        mid = (l + r) >> 1;

                        if(mp1[pa][mid] >= mp2[pb][b]){

                            r = mid;

                            a = min(a, mid);

                        }

                        else l = mid;

                    }

                }

            }

        }

        sza = cnt.size();

        for(auto k = cnt.begin(); k != cnt.end(); k++){

            ans[k->first] = sza;

        }

    }



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

        if(i == 1) cout << ans[i];

        else cout << " " << ans[i];

    }





    cout << endl;



    #ifdef LOCAL

    memset(ans, 0, sizeof ans);

    mp1.clear();

    mp2.clear();

    cout << endl;

    }

    #endif // LOCAL

    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: