Problem:
Score : 400 points

Problem Statement

An SNS has N users - User 1, User 2, \cdots, User N.

Between these N users, there are some relationships - M friendships and K blockships.

For each i = 1, 2, \cdots, M, there is a bidirectional friendship between User A_i and User B_i.

For each i = 1, 2, \cdots, K, there is a bidirectional blockship between User C_i and User D_i.

We define User a to be a friend candidate for User b when all of the following four conditions are satisfied:

a \neq b.

There is not a friendship between User a and User b.

There is not a blockship between User a and User b.

There exists a sequence c_0, c_1, c_2, \cdots, c_L consisting of integers between 1 and N (inclusive) such that c_0 = a, c_L = b, and there is a friendship between User c_i and c_{i+1} for each i = 0, 1, \cdots, L - 1.

For each user i = 1, 2, ... N, how many friend candidates does it have?

Constraints

All values in input are integers.

2 ≤ N ≤ 10^5

0 \leq M \leq 10^5

0 \leq K \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

1 \leq C_i, D_i \leq N

C_i \neq D_i

(A_i, B_i) \neq (A_j, B_j) (i \neq j)

(A_i, B_i) \neq (B_j, A_j)

(C_i, D_i) \neq (C_j, D_j) (i \neq j)

(C_i, D_i) \neq (D_j, C_j)

(A_i, B_i) \neq (C_j, D_j)

(A_i, B_i) \neq (D_j, C_j)

Input

Input is given from Standard Input in the following format:

N M K
A_1 B_1
\vdots
A_M B_M
C_1 D_1
\vdots
C_K D_K

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User 2 and 3, and between 3 and 4. Also, there is no friendship or blockship between User 2 and 4. Thus, User 4 is a friend candidate for User 2.

However, neither User 1 or 3 is a friend candidate for User 2, so User 2 has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main{
    static int N;
    static Map<Integer,Set<Integer>> friend;
    static Map<Integer, Set<Integer>> block;
    public static void main(String args[]){
        Scanner scanner = new Scanner(System.in);
        N = scanner.nextInt();
        int M = scanner.nextInt();
        int K = scanner.nextInt();
        friend = new HashMap<>();
        block = new HashMap<>();
        for(int i=0; i<M; i++){
            int A = scanner.nextInt();
            int B = scanner.nextInt();
            friend.putIfAbsent(A, new HashSet<>());
            friend.get(A).add(B);
            friend.putIfAbsent(B, new HashSet<>());
            friend.get(B).add(A);
        }
        for(int i=0; i<K; i++){
            int C = scanner.nextInt();
            int D = scanner.nextInt();
            block.putIfAbsent(C, new HashSet<>());
            block.get(C).add(D);
            block.putIfAbsent(D, new HashSet<>());
            block.get(D).add(C);
        }

        for(int i=1; i<=N; i++){
            Set<Integer> candidates = new HashSet<>();
            candidates.add(i);
            extend(i,candidates);
            if(friend.containsKey(i)){
                candidates.removeAll(friend.get(i));
            }
            if(block.containsKey(i)){
                candidates.removeAll(block.get(i));
            }
            candidates.remove(i);
            System.out.print(candidates.size());
            System.out.print(" ");
        }
    }
    public static void extend(int fr,Set<Integer> curset){
        if(friend.containsKey(fr)){
            for(int to :friend.get(fr)){
                if(!curset.contains(to)){
                    curset.add(to);
                    // System.out.println("extend "+String.valueOf(fr)+" to "+String.valueOf(to));
                    extend(to, curset);
                }
            }
        }
    }
}