stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 6.521387057906301x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <bits/stdc++.h>



using namespace std;



int fa[200005];

int fa1[200005];

int vis[200005];



int find1(int a)

{

    if(fa[a]!=a)

        fa[a]=find1(fa[a]);

    return fa[a];

}

void union1(int a,int b)

{

    fa[find1(a)]=find1(b);

}



int find2(int a)

{

    if(fa1[a]!=a)

        fa1[a]=find2(fa1[a]);

    return fa1[a];

}

void union2(int a,int b)

{

    fa1[find2(a)]=find2(b);

}

map<long long,int> mp;

int main()

{

    int n,k,l;

    cin>>n>>k>>l;

    memset(vis,0,sizeof(vis));

    for(int i=0;i<=n;i++)

    {

        fa[i]=i;

        fa1[i]=i;

    }

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

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        union1(l1,r);

    }

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

    {

        int l1,r;

        scanf("%d %d",&l1,&r);

        //if(find1(l1)==find1(r))

        union2(l1,r);

        //cout<<find1(l1)<<' '<<find1(r);

    }

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

    {

        if(mp.find((long long)((long long)find1(i)*200005+(long long)find2(i)))==mp.end())

            mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]=0;

        mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]++;

    }

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

        printf("%d ",mp[(long long)((long long)find1(i)*200005+(long long)find2(i))]);

    printf("%d\n",mp[(long long)((long long)find1(n)*200005+(long long)find2(n))]);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8191644531746594x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <cstdio>

#include <cstring>

#include <algorithm>

#include <vector>

#define ll long long

using namespace std;



const int maxn = 200010;

int n, m1, m2, u, v;

int f1[maxn], f2[maxn], vis[maxn], cnt[maxn], res[maxn];

vector <int> G[maxn];



int Find1 (int x) {

	if (f1[x] == x) return x;

	return f1[x] = Find1(f1[x]);

}



int Find2 (int x) {

	if (f2[x] == x) return x;

	return f2[x] = Find2(f2[x]);

}



int main () {

	scanf ("%d%d%d", &n, &m1, &m2);

	for (int i = 1; i <= n; i++) f1[i]= f2[i] = i;

	for (int i = 0; i < m1; i++) {

        scanf ("%d%d", &u, &v);

        f1[Find1(u)] = Find1(v);

	}

	for (int i = 0; i < m2; i++) {

        scanf ("%d%d", &u, &v);

        f2[Find2(u)] = Find2(v);

	}

    for (int i = 1; i <= n; i++) G[Find1(i)].push_back(i);

    for (int i = 1; i <= n; i++) if(G[i].size()) {

        for (int j = 0; j < G[i].size(); j++) {

            int t = Find2(G[i][j]);

            if (vis[t] != i) {

                vis[t] = i;

                cnt[t] = 0;

            }

            cnt[t]++;

        }

        for (int j = 0; j < G[i].size(); j++) {

            int t = Find2(G[i][j]);

            res[G[i][j]] = cnt[t];

        }

    }

    for (int i = 1; i <= n; i++) printf("%d%c", res[i], i==n?'\n':' ');

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 6.423885688185072x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



const int N = 2e5+10;

int fa1[N],fa2[N];

int findroot1(int u)

{

    return fa1[u]==u?u:fa1[u]=findroot1(fa1[u]);

}

int findroot2(int u)

{

    return fa2[u]==u?u:fa2[u]=findroot2(fa2[u]);

}

set<int>s2[N];

int ans[N];

int main()

{

    int n,k,l;

    int u,v;

    // freopen("in.txt","r",stdin);

    scanf("%d%d%d",&n,&k,&l);

    for(int i=1;i<=n;i++) fa1[i]=fa2[i]=i;

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

    {

        scanf("%d%d",&u,&v);

        int uu=findroot1(u);

        int vv=findroot1(v);

        if(uu==vv) continue;

        else fa1[vv]=uu;

    }

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

    {

        scanf("%d%d",&u,&v);

        int uu=findroot2(u);

        int vv=findroot2(v);

        if(uu==vv) continue;

        else fa2[vv]=uu;

    }

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

    {

        v=findroot2(i);

        s2[v].insert(i);

    }

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

    {

        if(!s2[i].empty())

        {

            map<int,int>mp;

            for(auto it:s2[i])

            {

                int uu=findroot1(it);

                mp[uu]++;

            }

            for(auto it:s2[i])

            {

                int uu=findroot1(it);

                ans[it]=mp[uu];

            }

        }

    }

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

    {

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

        if(i!=n) printf(" ");

        else printf("\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 12.107927563966262x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



#define X first

#define Y second

#define LL long long

#define ULL unsigned long long

#define PB push_back

#define MP make_pair

#define VI vector<int>

#define PII pair<int,int>

#define IOS ios::sync_with_stdio(0);cin.tie(0);

#define IN freopen("in", "r", stdin);

#define OUT freopen("out", "w", stdout);



const int maxn=200010;



int fa1[maxn],fa2[maxn];

int ans[maxn];

VI v[maxn];



int find1(int x)

{

	if (fa1[x]!=x) fa1[x]=find1(fa1[x]);

	return fa1[x];

}



int find2(int x)

{

	if (fa2[x]!=x) fa2[x]=find2(fa2[x]);

	return fa2[x];

}



bool cmp(int a,int b)

{

	return find1(a)<find1(b);

}



int main()

{

	IOS;

	int n,m1,m2;

	cin>>n>>m1>>m2;

	for (int j=1;j<=n;++j)

	{

		fa1[j]=j;

		fa2[j]=j;

	}

	for (int j=1;j<=m1;++j)

	{

		int x,y;

		cin>>x>>y;

		int fx=find1(x);

		int fy=find1(y);

		if (fx!=fy) fa1[fx]=fy;

	}

	for (int j=1;j<=m2;++j)

	{

		int x,y;

		cin>>x>>y;

		int fx=find2(x);

		int fy=find2(y);

		if (fx!=fy) fa2[fx]=fy;

	}

	for (int j=1;j<=n;++j)

	{

		int fj=find2(j);

		v[fj].PB(j);

	}

	for (int j=1;j<=n;++j)

	{

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

		sort(v[j].begin(),v[j].end(),cmp);

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

		{

			int p=k;

			while (p<v[j].size()-1&&find1(v[j][p+1])==find1(v[j][k])) ++p;

			int add=p-k+1;

			for (int l=k;l<=p;++l)

				ans[v[j][l]]+=add;

			k=p;

		}

	}

	for (int j=1;j<=n;++j)

		cout<<ans[j]<<" ";

	cout<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 6.548818889665158x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <queue>

#include <vector>

#include <string>

#include <set>

#include <map>

#define ll long long

#define ull unsigned long long

#define rep(i, to) for(int i = 0; i < (to); ++i)

#define rep1(i, to) for(int i = 1; i <= (to); ++i)

#define ms(x, v) memset(x, v, sizeof(x))

using namespace std;

const int N = 200005;

int pa1[N], pa2[N];

int fpa1(int x){ return pa1[x] == x ? x : pa1[x] = fpa1(pa1[x]); }

int fpa2(int x){ return pa2[x] == x ? x : pa2[x] = fpa2(pa2[x]); }

void lnk1(int x, int y) {x = fpa1(x), y = fpa1(y); if(x != y) pa1[x] = y;}

void lnk2(int x, int y) {x = fpa2(x), y = fpa2(y); if(x != y) pa2[x] = y;}

int ans[N];

vector<int> v[N];

int main()

{

#ifdef LOCAL

    freopen("in", "r", stdin);

#endif

//    ios::sync_with_stdio(0);

    int n, k, l, a, b;

    while(scanf("%d%d%d", &n, &k, &l) > 0)

    {

        ms(ans, 0);

        rep1(i, n) pa1[i] = pa2[i] = i, v[i].clear();

        rep1(i, k)

        {

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

            lnk1(a, b);

        }

        rep1(i, l)

        {

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

            lnk2(a, b);

        }

        rep1(i, n) v[fpa1(i)].push_back(i);

        rep1(i, n)

        {

            if(v[i].size())

            {

                map<int, int> mp;

                rep(j, (int)v[i].size()) mp[fpa2(v[i][j])]++;

                rep(j, (int)v[i].size())

                {

                    ans[v[i][j]] = mp[fpa2(v[i][j])];

                }

            }

        }

        rep1(i, n) printf("%d ", ans[i]);

        puts("");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.4484036981910675x):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <map>

#define rep(a,b,c) for(int a=b;a<=c;a++)

#define per(a,b,c) for(int a=b;a>=c;a--)

#define MAXN 200005

#define ll long long

using namespace std;

int f1[MAXN],f2[MAXN],res[MAXN];

typedef pair<int,int> pii;

map<pii,int> m;

void init(int n)

{

    rep(i,1,n)

        f1[i] = f2[i] = i,res[i] =1;

}

int getf(int x,int f[])

{

    if(x != f[x])

        f[x] = getf(f[x],f);

    return f[x];

}

void Union(int a,int b,int f[])

{

    a = getf(a,f);

    b = getf(b,f);

    if(a != b){

        f[a] = f[b];

    }

}

bool isSame(int a,int b,int f[])

{

    return getf(a,f) == getf(b,f);

}

int main()

{

    //freopen("data","r",stdin);

    int n,k,l;

    scanf("%d%d%d",&n,&k,&l);

    init(n);

    int a,b;

    rep(i,1,k)

    {

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

        Union(a,b,f1);

    }

    rep(i,1,l)

    {

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

        Union(a,b,f2);

    }

    rep(i,1,n){

        m[make_pair(getf(i,f1),getf(i,f2))]++;

    }

    rep(i,1,n)

    {

        printf("%d ",m[make_pair(getf(i,f1),getf(i,f2))]);

    }

    return 0;

}
// Slower C++ solution:
#include <cstdio>

#include <map>

#include <vector>

using namespace std;



const int N = 2e5 + 5;

int pa1[N], pa2[N];

int find(int* p, int v) { return v == p[v] ? v : p[v] = find(p, p[v]); }

void unite(int* p, int a, int b) { p[find(p, b)] = find(p, a); }



typedef long long LL;

int n, k, l;

// vector<int> e1[N], e2[N];



map<LL, int> mp;



int main() {

#ifdef D

    freopen("a.in", "r", stdin);

#endif

    for (int i = 0; i < N; i++) {

        pa1[i] = pa2[i] = i;

    }

    scanf("%d%d%d", &n, &k, &l);

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

        int a, b;

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

        unite(pa1, a, b);

        // e1[a].push_back(b);

        // e1[b].push_back(a);

    }

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

        int a, b;

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

        unite(pa2, a, b);

        // e2[a].push_back(b);

        // e2[b].push_back(a);

    }



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

        mp[find(pa1, i) * (LL)(n + 1) + find(pa2, i)] += 1;

    }



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

        printf("%d ", mp[find(pa1, i) * (LL)(n + 1) + find(pa2, i)]);

    }

    puts("");



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.5177892638476753x):
#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

#define inf (0x3f3f3f3f)

#define ls int len=s.size()

#define ls1 int len1=s1.size()

#define sd(x) scanf("%d",&x)

#define slld(x) scanf("%I64d",&x)

#define reps(i,n) for(int i=1;i<n;i++)

#define rep(i,n) for(int i=0;i<n;i++)

#define repn(i,n) for(int i=n-1;i>=0;i--)

typedef long long int LL;

typedef long long int ll;



#include <iostream>

#include <sstream>

#include <vector>

#include <set>

#include <map>

#include <queue>

#include <string>

#include<stack>

#include<ctime>

void g(int t){if(t){g((t-1)/26);putchar(65+(t-1)%26);}}



const int M=2*1e5+100;

int f[M],f1[M],r[M],r1[M];



void init(int n){

    rep(i,n){

        f[i]=i;f1[i]=i;

        r[i]=r1[i]=0;

    }

}

int find(int x){

    return f[x]==x?x:f[x]=find(f[x]);

}

int find1(int x){

    return f1[x]==x?x:f1[x]=find1(f1[x]);

}

void unite(int x,int y){

    x=find(x);y=find(y);

    if(x==y) return;



    if(r[x]<r[y]) f[x]=y;

    else{

        f[y]=x;

        if(r[x]==r[y]) r[x]++;

    }

}

void unite1(int x,int y){

    x=find1(x);y=find1(y);

    if(x==y) return;

    if(r1[x]<r1[y]) f1[x]=y;

    else{

        f1[y]=x;

        if(r1[x]==r1[y]) r1[x]++;

    }

}



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



int main()

{

#ifdef LOCAL

    freopen("data.h", "r", stdin);

    //freopen("1.h","w",stdout);

#endif

    int n,K,L;

    scanf("%d%d%d",&n,&K,&L);

    init(n);

    rep(i,K){

        int v,u;

        scanf("%d%d",&v,&u);

        unite(v,u);

    }

    rep(i,L){

        int v,u;

        scanf("%d%d",&v,&u);

        unite1(v,u);

    }

    mp.clear();

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

        mp[make_pair(find(i),find1(i))]++;

    }



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

        printf("%d ",mp[make_pair(find(i),find1(i))]);

    }

    cout<<endl;

}
// Slower C++ solution:
#include <cstdio>

#include <map>

#include <vector>

using namespace std;



const int N = 2e5 + 5;

int pa1[N], pa2[N];

int find(int* p, int v) { return v == p[v] ? v : p[v] = find(p, p[v]); }

void unite(int* p, int a, int b) { p[find(p, b)] = find(p, a); }



typedef long long LL;

int n, k, l;

// vector<int> e1[N], e2[N];



map<LL, int> mp;



int main() {

#ifdef D

    freopen("a.in", "r", stdin);

#endif

    for (int i = 0; i < N; i++) {

        pa1[i] = pa2[i] = i;

    }

    scanf("%d%d%d", &n, &k, &l);

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

        int a, b;

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

        unite(pa1, a, b);

        // e1[a].push_back(b);

        // e1[b].push_back(a);

    }

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

        int a, b;

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

        unite(pa2, a, b);

        // e2[a].push_back(b);

        // e2[b].push_back(a);

    }



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

        mp[find(pa1, i) * (LL)(n + 1) + find(pa2, i)] += 1;

    }



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

        printf("%d ", mp[find(pa1, i) * (LL)(n + 1) + find(pa2, i)]);

    }

    puts("");



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.49432953723759x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <map>

#include <vector>

#include <algorithm>

#include <cstdlib>

#include <cmath>

#include <bitset>

#define pb push_back

#define mp make_pair

#define x first

#define y second

#define pii pair<int,int>

#define pdd pair<db,db>



using namespace std;



typedef long long ll;

typedef double db;



const int N = 1e5+10,M=1e4+10,MOD = 1e9+7;



int fa[N*2],fb[2*N];

//int ans[N*2];



void init(int n){

    for(int i=0;i<=n;i++)

        fa[i]=fb[i]=i;

}



int finda(int x){

    if(fa[x]!=x)fa[x]=finda(fa[x]);

    return fa[x];

}

int findb(int x){

    if(fb[x]!=x)fb[x]=findb(fb[x]);

    return fb[x];

}



int main()

{

    int n,k,l;

    while(~scanf("%d%d%d",&n,&k,&l)){

        init(n);

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

            int x,y;

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

            int fx=finda(x),fy=finda(y);

            if(fx!=fy)fa[fx]=fy;

        }

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

            int x,y;

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

            int fx=findb(x),fy=findb(y);

            if(fx!=fy)fb[fx]=fy;

        }

        map<int,pii> m1;

        map<pii,int> m2;

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

            int xa=finda(i),xb=findb(i);

           //cout<<xa<<" "<<xb<<endl;

            m1[i]=mp(xa,xb);

            m2[mp(xa,xb)]++;

           // cout<<"("<<m1[i].x<<","<<m1[i].y<<") "<<m2[mp(xa,xb)]<<endl<<endl;

        }



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

            cout<<m2[m1[i]]<<" ";

        }

        cout<<endl;

       

    }

}