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

using namespace std;



long long f(long long a[],long long x)

{

    if(x==a[x]);

    else a[x]=f(a,a[x]);

    return a[x];

}

long long a[200010];

long long b[200010];

long long c[200010];

long long d[200010];

int e[200010];

int main()

{



    long long n,m,k;

    long long x,y,i;

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

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

    {

        a[i]=i;

        b[i]=i;

    }

    while(m--)

    {

        scanf("%lld%lld",&x,&y);

        x=f(a,x);

        y=f(a,y);

        a[x]=y;

    }

    while(k--)

    {

        scanf("%lld%lld",&x,&y);

        x=f(b,x);

        y=f(b,y);

        b[x]=y;

    }

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

    {

        c[i]=f(a,i)+(n+1)*f(b,i);

        d[i]=c[i];

    }

    sort(d+1,d+n+1);

    memset(e,0,sizeof(e));

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

    {

        y=lower_bound(d+1,d+n+1,c[i])-d;

        e[y]++;

    }

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

    {

        y=lower_bound(d+1,d+n+1,c[i])-d;

        if(i==n)printf("%d\n",e[y]);

        else printf("%d ",e[y]);

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7502157967274485x):
#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



//并查集可以直接写成1个类，然后2个并查集建2个对象就可以，这里就直接朴素写法的并查集了 Y ^_^ Y

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;

}