// 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: 9.601828775273479x):
#include<bits/stdc++.h>

using namespace std;

int bcj1[200010],bcj2[200010];

int father1(int k)

{

    return bcj1[k]==k?k:bcj1[k]=father1(bcj1[k]);

}

void join1(int x,int y)

{

    int fa,fb;

    fa=father1(x);fb=father1(y);

    bcj1[fb]=fa;

}

int father2(int k)

{

    return bcj2[k]==k?k:bcj2[k]=father2(bcj2[k]);

}

void join2(int x,int y)

{

    int fa,fb;

    fa=father2(x);fb=father2(y);

    bcj2[fb]=fa;

}

struct haha

{

    int p,q,no,ans;

}pp[200010];

bool cmp1(haha a,haha b)

{

    if (a.p==b.p) return a.q<b.q;

    return a.p<b.p;

}

bool cmp2(haha a,haha b)

{

    return a.no<b.no;

}

int main()

{

    int n,k,l,a,b,i,j,m,cnt;

    cin>>n>>k>>l;

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

    {

        bcj1[i]=bcj2[i]=i;

        pp[i].ans=0;pp[i].no=i;

    }

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

    {

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

        join1(a,b);

    }

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

    {

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

        join2(a,b);

    }

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

    {

        bcj1[i]=father1(bcj1[i]);

        pp[i].p=bcj1[i];

        bcj2[i]=father2(bcj2[i]);

        pp[i].q=bcj2[i];

    }

    sort(pp+1,pp+1+n,cmp1);cnt=0;

    pp[0].p=pp[0].q=0;j=1;

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

    {

        if (i<=n && pp[i].p==pp[i-1].p && pp[i].q==pp[i-1].q) continue;

        for(m=j;m<=i-1;m++)

            pp[m].ans=i-j;

        j=i;

    }

    sort(pp+1,pp+1+n,cmp2);

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

        printf("%d ",pp[i].ans);

    cout<<endl;

    return 0;

}