// 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: 5.3483278630534015x):
#include <iostream>

#include <bits/stdc++.h>



using namespace std;



int fa[200005];

int fa1[200005];

int vis[200005];



int find1(int a)

{

    if(fa[a]!=a)

        fa[a]=find1(fa[a]);

    return fa[a];

}

void union1(int a,int b)

{

    fa[find1(a)]=find1(b);

}



int find2(int a)

{

    if(fa1[a]!=a)

        fa1[a]=find2(fa1[a]);

    return fa1[a];

}

void union2(int a,int b)

{

    fa1[find2(a)]=find2(b);

}

map<long long,int> mp;

int main()

{

    int n,k,l;

    cin>>n>>k>>l;

    memset(vis,0,sizeof(vis));

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

    {

        fa[i]=i;

        fa1[i]=i;

    }

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

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        union1(l1,r);

    }

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

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        //if(find1(l1)==find1(r))

        union2(l1,r);

        //cout<<find1(l1)<<' '<<find1(r);

    }

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

    {

        if(mp.find((long long)((long long)find1(i)*200005+(long long)find2(i)))==mp.end())

            mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]=0;

        mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]++;

    }

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

        printf("%d ",mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]);

    printf("%d\n",mp[(long long)((long long)find1(n)*200005+(long long)find2(n))]);

    return 0;

}