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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
import java.util.StringJoiner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        Map<Integer, Set<Integer>> friends = new HashMap<>();
        Map<Integer, Set<Integer>> blocks = new HashMap<>();
        for (int i = 1; i <= n; i++) {
            friends.put(i, new HashSet<>());
            blocks.put(i, new HashSet<>());
        }
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            friends.get(a).add(b);
            friends.get(b).add(a);
        }
        for (int i = 0; i < k; i++) {
            int c = sc.nextInt();
            int d = sc.nextInt();
            blocks.get(c).add(d);
            blocks.get(d).add(c);
        }

        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            boolean[] visited = new boolean[n + 1];
            visited[i] = true;
            List<Integer> friendList = new ArrayList<>(friends.get(i));
            Set<Integer> friendSet = friends.get(i);
            Set<Integer> blockSet = blocks.get(i);
            Deque<Integer> queue = new ArrayDeque<>();
            for (int j = 0; j < friendList.size(); j++) {
                queue.add(friendList.get(j));
                visited[friendList.get(j)] = true;
            }
            int cnt = 0;
            while (!queue.isEmpty()) {
                int size = queue.size();
                for (int j = 0; j < size; j++) {
                    int f = queue.poll();
                    List<Integer> li = new ArrayList<>(friends.get(f));
                    for (int l = 0; l < li.size(); l++) {
                        int ff = li.get(l);
                        if (visited[ff]) {
                            continue;
                        }
                        if (!friendSet.contains(ff) && !blockSet.contains(ff)) {
                            cnt++;
                        }
                        visited[ff] = true;
                        queue.add(ff);
                    }
                }
            }
            ans.add(cnt);
        }

        StringJoiner sj = new StringJoiner(" ");
        for (int i = 0; i < n; i++) {
            sj.add(String.valueOf(ans.get(i)));
        }
        System.out.println(sj);
    }

}