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

//AtCoder Beginner Contest 157
//D - Friend Suggestions
public class Main {

	static void dfs(List<List<Integer>> g, int i, List<Boolean> seen) {
		seen.set(i, true);

		List<Integer> vList = g.get(i);
		for (Integer v : vList) {
			if (seen.get(v))
				continue;
			dfs(g, v, seen);
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		int M = Integer.parseInt(sc.next());
		int K = Integer.parseInt(sc.next());
		List<List<Integer>> g = new ArrayList<>();
		for (int i = 0; i < N; i++) {
			g.add(new ArrayList<>());
		}
		for (int i = 0; i < M; i++) {
			int A = Integer.valueOf(sc.next()) - 1;
			int B = Integer.valueOf(sc.next()) - 1;
			g.get(A).add(B);
			g.get(B).add(A);
		}

		List<List<Integer>> block = new ArrayList<>();
		for (int i = 0; i < N; i++) {
			block.add(new ArrayList<>());
		}
		for (int i = 0; i < K; i++) {
			int C = Integer.valueOf(sc.next()) - 1;
			int D = Integer.valueOf(sc.next()) - 1;
			block.get(C).add(D);
			block.get(D).add(C);
		}
		sc.close();

		List<Integer> ans = new ArrayList<>();
		for (int i = 0; i < N; i++) {
			List<Boolean> seen = new ArrayList<>(N);
			for (int b = 0; b < N; b++) {
				seen.add(false);
			}
			dfs(g, i, seen);

			seen.set(i, false);
			for (int f = 0; f < g.get(i).size(); f++) {
				seen.set(g.get(i).get(f), false);
			}
			for (int k = 0; k < block.get(i).size(); k++) {
				seen.set(block.get(i).get(k), false);
			}
			int a = 0;
			for (int b = 0; b < N; b++) {
				if (seen.get(b)) {
					a++;
				}
			}
			ans.add(a);
		}
		for (Integer i : ans) {
			System.out.print(i + " ");
		}
	}
}