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: Runtime Error)

import java.util.*;

public class Main {

	static class Unionfind{
		//基本的にマイナスのサイズを持つ
		//根でなければ非負のおやのIDを持つ
		private int d[];
		
		public Unionfind(int n) {
			d = new int[n + 1];
			//初めは全員根でサイズは１なので-1で初期化
			for (int i = 0; i <= n; ++i)
				d[i] = -1;
		}
		
		public int root(int x) {
			if (d[x] < 0)
			return x;
			return d[x] = root(d[x]);
		}
		
		public boolean unite(int x, int y) {
			int rx = root(x);
			int ry = root(y);
			if(rx == ry)return false;
			//もしxのグループのほうがサイズが小さければxの親をyとしなければいけないので入れ替える。
			if(d[x] > d[y]){
				int tmp = x;
				x = y;
				y = tmp;
			}
			//yの親をxにするのでIDxのグループのサイズがd[y]だけ増える
			d[x] += d[y];
			d[y] = x;
			return true;
		}
		
		public boolean isSame(int x, int y) {
			int rx = root(x);
			int ry = root(y);
			return rx == ry;
		}
		
		public int size(int x){
			int rx = root(x);
			return -d[rx];
		}
	}
	
	static Unionfind uniontree;
	
	public static class Graph {
		List<Integer> friend = new ArrayList<Integer>();
		List<Integer> block = new ArrayList<Integer>();
		int number;

		public Graph(int num){
			number = num;
		}

		public void addfriend(int b) {
			friend.add(b);
		}

		public void addblock(int b) {
			if(uniontree.isSame(number, b))
			block.add(b);
		}
	}

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
		Graph graph[] = new Graph[n + 1];
		uniontree = new Unionfind(n);
		for (int i = 0; i <= n; ++i)
			graph[i] = new Graph(i);
		for (int i = 0; i < m; ++i) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			graph[a].addfriend(b);
			graph[b].addfriend(a);
			uniontree.unite(a, b);
		}
		for (int i = 0; i < k; ++i) {
			int c = sc.nextInt();
			int d = sc.nextInt();
			graph[c].addblock(d);
			graph[d].addblock(c);
		}
		sc.close();
		for (int i = 1; i <= n; ++i) {
			int num = uniontree.size(i) - (graph[i].friend.size() + graph[i].block.size()) - 1;
			// int num = uniontree.size(i);
			System.out.print(num + " ");
		}
	}
}