// 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: 2.285891653404446x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#include <string>

#include <stack>

#include <map>

#include <ctime>

#include <set>

#include <bitset>

#include <ctime>

#define X first

#define Y second

#define clr(u,v); memset(u,v,sizeof(u));

#define in() freopen("data","r",stdin);

#define out() freopen("data","w",stdout);

#define Clear(Q); while (!Q.empty()) Q.pop();

#define pb push_back

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<char, int> pci;

const int maxn = 2e5 + 10;

const int INF = 0x3f3f3f3f;

int f[maxn];

int f2[maxn];

int f3[maxn];



int sum[maxn];

void init(int n)

{

    for (int i = 0; i < maxn; i++)

        f[i] = i, f2[i] = i, f3[i] = i, sum[i] = 1;

}

int find(int x)

{

    return f[x] == x ? x : f[x] = find(f[x]);

}

int find2(int x)

{

    return f2[x] == x ? x : f2[x] = find2(f2[x]);

}



int find3(int x)

{

    return f3[x] == x ? x : f3[x] = find3(f3[x]);

}





void mix(int x, int y)

{

    int fx = find(x), fy = find(y);

    if (fx == fy) return ;

    f[fx] = fy;

}

void mix2(int x, int y)

{

    int fx = find2(x), fy = find2(y);

    if (fx == fy) return ;

    f2[fx] = fy;

}

void mix3(int x, int y)

{

    int fx = find3(x), fy = find3(y);

    if (fx == fy) return ;

    f3[fx] = fy;

    sum[fy] += sum[fx];

}

map <pii, int> M;



int main()

{

#ifdef LOCAL

//    in();

#endif

    int n, k, l, a, b;

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

    init(n);

    while (k--)

    {

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

        mix(a, b);

    }

    while (l--)

    {

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

        mix2(a, b);

//        else mix3(a,b);

    }

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

        M[make_pair(find(i), find2(i))]++;



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

        printf("%d ", M[make_pair(find(i), find2(i))]);

    puts("");

    return 0;

}