// Original C++ solution (slower):
#include<cstdio>

#include<algorithm>

#include<iostream>

#include<cstring>

#include<vector>

#include<cmath>

#include<map>

#include<set>

#include<queue>

using namespace std;

typedef long long ll;

const int N=2e5+10;

int a[N],b[N];

vector<int>v[N];

map<pair<int,int>,int>mp;

queue<int>q;

int main()

{

    int n,k,l,i,j;

    cin>>n>>k>>l;

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

    {

        int x,y;

        scanf("%d %d",&x,&y);

        v[x].push_back(y);

        v[y].push_back(x);

    }

    int cnt=0;

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

    if(a[i]==0)

    {

        a[i]=++cnt;

        q.push(i);

        while(!q.empty())

        {

            int x=q.front();

            for(j=0;j<v[x].size();j++)

            if(a[v[x][j]]==0)

            {

                a[v[x][j]]=cnt;

                q.push(v[x][j]);

            }

            q.pop();

        }

    }

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

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

    {

        int x,y;

        scanf("%d %d",&x,&y);

        v[x].push_back(y);

        v[y].push_back(x);

    }

    cnt=0;

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

    if(b[i]==0)

    {

        b[i]=++cnt;

        q.push(i);

        while(!q.empty())

        {

            int x=q.front();

            for(j=0;j<v[x].size();j++)

            if(b[v[x][j]]==0)

            {

                b[v[x][j]]=cnt;

                q.push(v[x][j]);

            }

            q.pop();

        }

    }

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

        mp[make_pair(a[i],b[i])]++;

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

        printf("%d ",mp[make_pair(a[i],b[i])]);

	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: