Problem:
Score : 400 points

Problem Statement

There are N cities. There are also K roads and L railways, extending between the cities.
The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities.
No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.

We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads.
We will also define connectivity by railways similarly.

For each city, find the number of the cities connected to that city by both roads and railways.

Constraints

2 ≦ N ≦ 2*10^5

1 ≦ K, L≦ 10^5

1 ≦ p_i, q_i, r_i, s_i ≦ N

p_i < q_i

r_i < s_i

When i ≠ j, (p_i, q_i) ≠ (p_j, q_j)

When i ≠ j, (r_i, s_i) ≠ (r_j, s_j)

Input

The input is given from Standard Input in the following format:

N K L
p_1 q_1
:
p_K q_K
r_1 s_1
:
r_L s_L

Output

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

Sample Input 1

4 3 1
1 2
2 3
3 4
2 3

Sample Output 1

1 2 2 1

All the four cities are connected to each other by roads.

By railways, only the second and third cities are connected. Thus, the answers for the cities are 1, 2, 2 and 1, respectively.

Sample Input 2

4 2 2
1 2
2 3
1 4
2 3

Sample Output 2

1 2 2 1

Sample Input 3

7 4 4
1 2
2 3
2 5
6 7
3 5
4 5
3 4
6 7

Sample Output 3

1 1 2 1 2 2 2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long int lli;
typedef unsigned long long int ulli;
typedef pair<int,int> pii;
typedef priority_queue< int, vector< int >, greater< int > >  minHeap;

#define ones(x) __builtin_popcount(x)
#define onesl(x) __builtin_popcountl(x)
#define onesll(x) __builtin_popcountll(x)

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define scn(n) scanf("%d",&n)
#define scnll(n) scanf("%lld",&n)
#define scn2(n,m) scanf("%d%d",&n,&m)
#define scn3(n,m,w) scanf("%d%d%d",&n,&m,&w)
#define scn2ll(n,m) scanf("%lld%lld",&n,&m)
#define atoz(v) v.begin(),v.end()
#define ratoz(v) v.rbegin(),v.rend()
#define Fill(a,v) memset(a,v,sizeof(a))
#define sz(v) v.size()
#define fi first
#define se second
#define inf 1e9
#define pi acos(-1.0)
#define sqr(x) x*x
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define ten(n) (int)1e##n
#define tenll(n) (lli)1e##n
#define mod 1000000007
#define FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

int Set(int N,int pos)
{
    return N=N | (1<<pos);
}
int reset(int N,int pos)
{
    return N = N & ~(1<<pos);
}
bool check(int N,int pos)
{
    return (bool)(N & (1<<pos));
}


inline int addmod(int x,int y)
{
    return (x%mod + y%mod)%mod;
}
inline int submod(int x,int y)
{
    return (x%mod - y%mod + mod)%mod;
}
inline int mulmod(int x,int y)
{
    return (x%mod *1LL* y%mod)%mod;
}
inline int nextSubMask(int i, int mask)
{
    return (i-1)&mask;   /// returns next smallest submask
}

void we_r_done()
{
    cout<<"NO";
    exit(0);
}

void debug(bool ok, string mssg)
{
    if(ok)
        return;
    cout<<mssg;
    exit(0);
}
#define N 300003

vector<int>road[N];
vector<int>rail[N];
bool posbyroad[N], both[N];
int cnt = 0;
int ans[N];
int vis[N];
int curr=1;
vector<int>conn;

void dfs(int x)
{

    posbyroad[x] = 1;
    conn.pb(x);

    for(int u : road[x])
    {

        if(ans[u] || posbyroad[u])
            continue;

        dfs(u);
    }
}
vector<int>now;

void dfs1(int x)
{

    if(posbyroad[x]){
        cnt++;
        now.pb(x);
    }
    both[x] = 1;
    // cout<<cnt<<" "<<x<<endl;

    for(int u : rail[x])
    {

        if(ans[u] || both[u])
            continue;

        dfs1(u);
    }
}

int main()
{
    ///freopen("output.txt","w",stdout);
    ///freopen("input.txt","r",stdin);
    int n,r,l;
    scn3(n,r,l);

    for(int i=0; i<r; i++)
    {
        int u,v;
        scn2(u,v);

        road[u].pb(v);
        road[v].pb(u);
    }

    for(int i=0; i<l; i++)
    {
        int u,v;
        scn2(u,v);

        rail[u].pb(v);
        rail[v].pb(u);
    }

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

        if(ans[i] || posbyroad[i])
            continue;
        //assert(posbyroad[i]==false && both[i]==false);
        dfs(i);

        while(conn.size())
        {
            int ii = conn.back();
            conn.pop_back();
            if(ans[ii]) continue;
            cnt = 0;
Fill(both,false);
            dfs1(ii);

            while(now.size())
            {
                ans[now.back()]  = cnt;
                now.pop_back();
            }
        }
    }

    for(int i=1; i<=n; i++)
    {
        printf("%d ",ans[i]);
    }

    return 0;

}

///sin and cos expect input in radians not degrees. so use , sin(degrees * pi / 180)
///using bs = bitset<MX>; // how many distinct number can be form?