// 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: 3.733301424666227x):
#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long LL;

inline int read() {

	int f=1,x=0; char ch=getchar();

	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }

	while (ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); }

	return x*f;

}



const int MAXN=200005;

int n,m,r,tot,v[MAXN],f[MAXN],ans[MAXN];

struct List{

	int l,lt[MAXN],nt[MAXN<<1],v[MAXN<<1];

	void clear() { l=0; memset(lt,0,sizeof(lt)); }

	void addedge(int x,int y) { nt[++l]=lt[x]; lt[x]=l; v[l]=y; }

}E;

struct Data{

	int x,y;

	bool operator <(const Data &B)const { return x<B.x; }

}A[MAXN];

int ask(int x) { if (f[x]==x) return x; f[x]=ask(f[x]); return f[x]; }

void dfs(int x) {

	v[x]=1; A[++tot].x=ask(x); A[tot].y=x;

	for (int i=E.lt[x];i;i=E.nt[i]) if (!v[E.v[i]]) dfs(E.v[i]);

}

int main() {

	int i,j,k,l;

	n=read(); m=read(); r=read();

	for (i=1;i<=n;++i) f[i]=i; E.clear();

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

		j=read(); k=read();

		E.addedge(j,k); E.addedge(k,j);

	}

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

		j=ask(read()); k=ask(read());

		if (j!=k) f[j]=k;

	}

	memset(v,0,sizeof(v));

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

	if (!v[i]) {

		tot=0; dfs(i); sort(A+1,A+1+tot);

		for (j=1;j<=tot;j=k+1) {

			for (k=j;k<tot&&A[k+1].x==A[j].x;++k);

			for (l=j;l<=k;++l) ans[A[l].y]=k-j+1;

		}

	}

	for (i=1;i<n;++i) printf("%d ",ans[i]); printf("%d\n",ans[n]);

	return 0;

}



/*

4 3 1

1 2

2 3

3 4

2 3



*/