stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



const int maxn = 200010;



int fa1[maxn], fa2[maxn], r1[maxn], r2[maxn];



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



void kexin_init() {

	for (int i=1; i<=maxn; ++i) {fa1[i] = i; fa2[i] = i;}

	memset(r1, 0, sizeof r1);

	memset(r2, 0, sizeof r2);

}



int kexin_find_road(int x) {

	if (fa1[x] == x) return x;

	return fa1[x] = kexin_find_road(fa1[x]);

}

void kexin_Union_road(int x, int y) {

	x = kexin_find_road(x);

	y = kexin_find_road(y);

	if (x == y) return ;

	if (r1[x] < r1[y]) swap(x, y);

	fa1[y] = x;

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

}



int kexin_find_railway(int x) {

	if (fa2[x] == x) return x;

	return fa2[x] = kexin_find_railway(fa2[x]);

}

void kexin_Union_railway(int x, int y) {

	x = kexin_find_railway(x);

	y = kexin_find_railway(y);

	if (x == y) return ;

	if (r2[x] < r2[y]) swap(x, y);

	fa2[y] = x;

	if (r2[x] == r2[y]) r2[x]++;

}

int main() {

	int n, k, l;

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

	

	kexin_init();

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

		int x, y;

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

		kexin_Union_road(x, y);

	}

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

		int x, y;

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

		kexin_Union_railway(x, y);

	}

	

	mp.clear();



	for (int i=1; i<=n; ++i) mp[make_pair(kexin_find_road(i), kexin_find_railway(i))]++;

	for (int i=1; i<n; ++i) printf("%d ", mp[make_pair(kexin_find_road(i), kexin_find_railway(i))]);

	printf("%d\n", mp[make_pair(kexin_find_road(n), kexin_find_railway(n))]);



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.159339094939843x):
#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;

       

    }

}
// Slower C++ solution:
#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



*/


// Faster C++ solution for the same problem (measured speedup: 2.5514824489085606x):
#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;

       

    }

}
// 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: 1.781015170615036x):
#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;

       

    }

}
// 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: 6.28917184754157x):
#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;

       

    }

}
// 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: 11.854014978414627x):
#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;

       

    }

}
// 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: 6.411485103366922x):
#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;

       

    }

}
// 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: 2.5277970634656644x):
#include <iostream>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <stack>

#include <queue>

#include <map>

#include <set>

#include <string>

#include <cstring>

#include <bitset>

#define ll long long

#define rep(x,n) for(ll (x)=0;(x)<n;(x)++)

#define rep1(x,n) for(ll (x)=1;(x)<=n;(x)++)

#define lson(x) ((x)<<1)

#define rson(x) ((x)<<1|1)

using namespace std;



const int N=200005;

int pa1[N],pa2[N],ans[N];

map<pair<int,int>,int>cnt;



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]);

}



int main()

{

    #ifdef LOCAL

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

    #endif // LOCAL

    int n,k,l,a,b;

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

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

    while(k--)

    {

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

        pa1[fpa1(a)]=fpa1(b);

    }

    while(l--)

    {

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

        pa2[fpa2(a)]=fpa2(b);

    }

    rep1(i,n) cnt[make_pair(fpa1(i),fpa2(i))]++;

    rep1(i,n) ans[i]=cnt[make_pair(fpa1(i),fpa2(i))];

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

    printf("\n");

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

using namespace std;



const int maxn = 200010;



int fa1[maxn], fa2[maxn], r1[maxn], r2[maxn];



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



void kexin_init() {

	for (int i=1; i<=maxn; ++i) {fa1[i] = i; fa2[i] = i;}

	memset(r1, 0, sizeof r1);

	memset(r2, 0, sizeof r2);

}



int kexin_find_road(int x) {

	if (fa1[x] == x) return x;

	return fa1[x] = kexin_find_road(fa1[x]);

}

void kexin_Union_road(int x, int y) {

	x = kexin_find_road(x);

	y = kexin_find_road(y);

	if (x == y) return ;

	if (r1[x] < r1[y]) swap(x, y);

	fa1[y] = x;

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

}



int kexin_find_railway(int x) {

	if (fa2[x] == x) return x;

	return fa2[x] = kexin_find_railway(fa2[x]);

}

void kexin_Union_railway(int x, int y) {

	x = kexin_find_railway(x);

	y = kexin_find_railway(y);

	if (x == y) return ;

	if (r2[x] < r2[y]) swap(x, y);

	fa2[y] = x;

	if (r2[x] == r2[y]) r2[x]++;

}

int main() {

	int n, k, l;

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

	

	kexin_init();

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

		int x, y;

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

		kexin_Union_road(x, y);

	}

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

		int x, y;

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

		kexin_Union_railway(x, y);

	}

	

	mp.clear();



	for (int i=1; i<=n; ++i) mp[make_pair(kexin_find_road(i), kexin_find_railway(i))]++;

	for (int i=1; i<n; ++i) printf("%d ", mp[make_pair(kexin_find_road(i), kexin_find_railway(i))]);

	printf("%d\n", mp[make_pair(kexin_find_road(n), kexin_find_railway(n))]);



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.215146793230389x):
#include <iostream>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <stack>

#include <queue>

#include <map>

#include <set>

#include <string>

#include <cstring>

#include <bitset>

#define ll long long

#define rep(x,n) for(ll (x)=0;(x)<n;(x)++)

#define rep1(x,n) for(ll (x)=1;(x)<=n;(x)++)

#define lson(x) ((x)<<1)

#define rson(x) ((x)<<1|1)

using namespace std;



const int N=200005;

int pa1[N],pa2[N],ans[N];

map<pair<int,int>,int>cnt;



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]);

}



int main()

{

    #ifdef LOCAL

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

    #endif // LOCAL

    int n,k,l,a,b;

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

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

    while(k--)

    {

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

        pa1[fpa1(a)]=fpa1(b);

    }

    while(l--)

    {

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

        pa2[fpa2(a)]=fpa2(b);

    }

    rep1(i,n) cnt[make_pair(fpa1(i),fpa2(i))]++;

    rep1(i,n) ans[i]=cnt[make_pair(fpa1(i),fpa2(i))];

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

    printf("\n");

}