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: Wrong Answer)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

class Person {
    List<Person> friends = new ArrayList<Person>();
    List<Person> blocks = new ArrayList<Person>();
    int name;
    int friendNum = -1;

    Person(int name) {
        this.name = name;
    }

    public void addFriends(Person p) {
        this.friends.add(p);
    }

    public void addBlocks(Person p) {
        this.blocks.add(p);
    }

    public List<Person> getFriends() {
        return this.friends;
    }

    public List<Person> getBlocks() {
        return this.blocks;
    }

    public List<Person> traverseFriends(List<Person> donePerson) {
        donePerson.add(this);
        for (Person person : this.friends) {
            if (donePerson.contains(person)) continue;
            donePerson = person.traverseFriends(donePerson);
        }

        return donePerson;
    }

    public int getFriendNum() {
        return this.friendNum;
    }

    public List<Person> calcFriendNum(List<Person> donePerson, List<Person> cand) {
        donePerson.add(this);

        int total = 0;
        for (Person person : cand) {
            if (this.equals(person) ||
                this.getFriends().contains(person) ||
                this.getBlocks().contains(person)) {

                } else {
                    total++;
            }
        }
        this.friendNum = total;

        for (Person person : this.friends) {
            if (donePerson.contains(person)) continue;
            donePerson = person.calcFriendNum(donePerson, cand);
        }
        return donePerson;
    }
}

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        int k = scan.nextInt();

        List<Person> ps = new ArrayList<Person>();
        for (int i = 0; i < n; i++) {
            ps.add(new Person(i));
        }

        for (int i = 0; i < m; i++) {
            int a = scan.nextInt();
            int b = scan.nextInt();
            ps.get(a - 1).addFriends(ps.get(b - 1));
            ps.get(b - 1).addFriends(ps.get(a - 1));
        }

        for (int i = 0; i < k; i++) {
            int a = scan.nextInt();
            int b = scan.nextInt();
            ps.get(a - 1).addBlocks(ps.get(b - 1));
            ps.get(b - 1).addBlocks(ps.get(a - 1));
        }

        for (int i = 0; i < n; i++) {
            int num = ps.get(i).getFriendNum();
            if (num == -1) {
                List<Person> cand = ps.get(i).traverseFriends(new ArrayList<Person>());
                ps.get(i).calcFriendNum(new ArrayList<Person>(), cand);
                num = ps.get(0).getFriendNum();
            }
            System.out.print(num + " ");
        }
    }
}