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: Accepted)

import java.util.*;
import java.util.stream.Collectors;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {
	int N, M, K;
	int[] A, B, C, D;
	Node[] nodes;
	Map<Integer, Set<Integer>> groupInfos;
	Set<Integer> reached;

	static class Node {
		Set<Integer> blocks;
		Set<Integer> friends;
		Set<Integer> kouhos;
		int id;
		int group;
		int kouhoNum;

		Node(int id) {
			this.id = id;
			this.group = -1;
			this.blocks = new HashSet<>();
			this.friends = new HashSet<>();
			this.kouhos = new HashSet<>();
		}

		void setKouhoNum(Set<Integer> ids) {
			this.kouhoNum = ids.size() - 1 - friends.size() - blocks.size();
		}
	}

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	Main(BufferedReader in) throws IOException {
		String[] tokens = in.readLine().split(" ");
		N = Integer.parseInt(tokens[0]);
		M = Integer.parseInt(tokens[1]);
		K = Integer.parseInt(tokens[2]);
		this.A = new int[M];
		this.B = new int[M];
		for (int i = 0; i < M; ++i) {
			tokens = in.readLine().split(" ");
			A[i] = Integer.parseInt(tokens[0]);
			B[i] = Integer.parseInt(tokens[1]);
		}
		C = new int[K];
		D = new int[K];
		for (int i = 0; i < K; ++i) {
			tokens = in.readLine().split(" ");
			C[i] = Integer.parseInt(tokens[0]);
			D[i] = Integer.parseInt(tokens[1]);
		}
		nodes = new Node[N];
		for (int i = 0; i < N; ++i) {
			nodes[i] = new Node(i);
		}
		for (int i = 0; i < M; ++i) {
			nodes[A[i] - 1].friends.add(B[i] - 1);
			nodes[B[i] - 1].friends.add(A[i] - 1);
		}
		for (int i = 0; i < K; ++i) {
			nodes[C[i] - 1].blocks.add(D[i] - 1);
			nodes[D[i] - 1].blocks.add(C[i] - 1);
		}
		this.reached = new HashSet<>();
	}

	void setGroup(Node node, int parent, int group) {
		if (node.group >= 0) {
			// すでにセット済み
			return;
		}
		node.group = group;
		for (Integer friend : node.friends) {
			if (nodes[friend].group < 0) {
				setGroup(nodes[friend], node.id, group);
			}
		}
	}

	void calc() {
		// まずグループをセットする
		int groupId = 0;
		for (int i = 0; i < N; ++i) {
			if (nodes[i].group < 0) {
				setGroup(nodes[i], -1, groupId);
				++groupId;
			}
		}

		// groupInfoを作成
		this.groupInfos = new HashMap<>();
		for (int i = 0; i < N; ++i) {
			Node node = nodes[i];
			if (!groupInfos.containsKey(node.group)) {
				groupInfos.put(node.group, new HashSet<>());
			}
			groupInfos.get(node.group).add(node.id);
		}
		// groupInfoをもとに値をセット
		for (int i = 0; i < N; ++i) {
			Node node = nodes[i];
			node.setKouhoNum(groupInfos.get(node.group));
		}

		// groupが異なるblock関係を削除する
		for (int i = 0; i < K; ++i) {
			int s = C[i] - 1;
			int t = D[i] - 1;
			if (nodes[s].group != nodes[t].group) {
				nodes[s].blocks.remove(t);
				nodes[t].blocks.remove(s);
			}
		}
		for (int i = 0; i < N; ++i) {
			Node node = nodes[i];
			node.setKouhoNum(groupInfos.get(node.group));
		}
	}

	void show() {
		for (int i = 0; i < N; ++i) {
			for (Integer id : nodes[i].kouhos) {
				System.out.print(id + " ");
			}
			System.out.println("");
		}
	}

	void showResult() {
		// show();
		StringBuilder strBuilder = new StringBuilder();
		strBuilder.append(Integer.toString(nodes[0].kouhoNum));
		for (int i = 1; i < N; ++i) {
			strBuilder.append(" " + Integer.toString(nodes[i].kouhoNum));
		}
		System.out.println(strBuilder.toString());
	}

}