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

}


// 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: