// Original C++ solution (slower):
#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



*/


// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

struct abc {

	int aa, bb, cc;

	bool operator < (const abc&T) const {

		return(aa < T.aa || (aa == T.aa && bb < T.bb) || (aa == T.aa && bb == T.bb && cc < T.cc));

	}

	bool operator == (const abc&T) const {

		return(aa == T.aa && bb == T.bb);

	}

} ans[200005];

using namespace std;

int N, KK, LL, i, j, k, l, m, n, pt1[200005], pt2[200005], s1[200005], s2[200005], z;

int sz[200005], ans2[200005];

int find1(int aa) {

	if(pt1[aa] != aa) 

		pt1[aa] = find1(pt1[aa]);

	return(pt1[aa]);

}

int find2(int bb) {

	if(pt2[bb] != bb) 

		pt2[bb] = find2(pt2[bb]);

	return(pt2[bb]);

}

int main() {

	scanf("%d%d%d", &N, &KK, &LL);

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

		pt1[i] = i;

		pt2[i] = i;

		s1[i] = 1;

		s2[i] = 1;

	}

	for(i = 0; i < KK; i++) {

		scanf("%d%d", &j, &l);

		if(find1(j) == find1(l)) continue;

		s1[find1(l)] += s1[find1(j)];

		pt1[find1(j)] = find1(l);

	}

	

	for(i = 0; i < LL; i++) {

		scanf("%d%d", &j, &l);

		if(find2(j) == find2(l)) continue;

		s2[find2(l)] += s2[find2(j)];

		pt2[find2(j)] = find2(l);

	}

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

		pt1[i] = find1(i);

		pt2[i] = find2(i);

		ans[i].aa = pt1[i];

		ans[i].bb = pt2[i];

		ans[i].cc = i;

	}

//	for(i = 1; i <= N; i++) printf("%d %d\n", pt1[i], pt2[i]);

	

	sort(ans + 1, ans + N + 1);

	sz[1] = 1;

	for(i = 2; i <= N; i++) if(ans[i] == ans[i - 1]) {

		sz[i] = sz[i - 1] + 1;

	} else {

		sz[i] = 1;

	}

	for(i = N - 1; i >= 1; i--) 

	if(ans[i] == ans[i + 1]) sz[i] = sz[i + 1];

	

	//for(i = 1; i <= N; i++) printf("%d %d %d %d\n", ans[i].aa, ans[i].bb, ans[i].cc, sz[i]);

	

	for(i = 1; i <= N; i++) ans2[ans[i].cc] = sz[i];

	

	for(i = 1; i < N; i++) printf("%d ", ans2[i]);

	printf("%d\n", ans2[N]);

}

// Fastest known solution for this problem: