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, M, K;
	
	//
	// Depth First Search により友達の友達を洗い出す
	//
	public static void dfs(List<List<Integer>> graph, List<Integer> known_list, int s){
   	
    	// 自分に到達可能
    	known_list.add(s);

    	for(int next_s: graph.get(s)){
    		if(known_list.contains(next_s)){
    			// 探索済み
    			continue;
    		}else{
    			dfs(graph, known_list, next_s);
    		}
    	}
	}
	
	public static void main(String[] args) {

		//
		// 画面入力
		//
		Scanner sc = new Scanner(System.in);
    	N = sc.nextInt();
    	M = sc.nextInt();
    	K = sc.nextInt();

    	//
    	// グラフ作成 i=0は使用しないのでN+1作成する
    	// Adjacency list representation 
    	// 0.05(s) 可読性悪
    	//
    	List<List<Integer>> graph_friends = new ArrayList<List<Integer>>();
    	List<List<Integer>> graph_block = new ArrayList<List<Integer>>();
    	List<List<Integer>> graph_seen = new ArrayList<List<Integer>>();
    	for (int i=0; i<N+1; i++) {
    		graph_friends.add(new ArrayList<Integer>());
    		graph_block.add(new ArrayList<Integer>());
    		graph_seen.add(new ArrayList<Integer>());
    	}
    	
    	// 友達入力
    	for (int i=0; i<M; i++) {
        	int a=sc.nextInt();
        	int b=sc.nextInt();
        	graph_friends.get(a).add(b);
        	graph_friends.get(b).add(a);
    	}

    	//System.out.println(graph_friends);

    	// ブロック入力
    	for (int i=0; i<K; i++) {
        	int a=sc.nextInt();
        	int b=sc.nextInt();
        	graph_block.get(a).add(b);
        	graph_block.get(b).add(a);
    	}
    	
    	//System.out.println(graph_block);

    	//
    	// 深さ優先探索 。まずはstart:1、end:3
    	//
    	for(int i=0; i<N+1; i++){
    		
    		List<Integer> known_list =  new ArrayList<Integer>();
        	dfs(graph_friends, known_list, i);
        	

        	// seenから友達とブロックを引く
        	known_list.removeAll(graph_friends.get(i));
        	known_list.removeAll(graph_block.get(i));
        	if(i != 0){
            	System.out.print(known_list.size() - 1);
            	if(i != N){
            		System.out.print(" ");
            	}
        	}
    	}
   	
    	sc.close();
	}

}