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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

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

		ArrayList<HashSet> setList = new ArrayList<HashSet>();
		int[] counts = new int[n];
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int index1 = getSetIndex(setList, a);
			int index2 = getSetIndex(setList, b);
			if (index1 == -1 && index2 == -1) {
				HashSet set = new HashSet();
				set.add(a);
				set.add(b);
				setList.add(set);
			} else if (index1 == -1) {
				setList.get(index2).add(a);
			} else if (index2 == -1) {
				setList.get(index1).add(b);
			} else if (index1 != index2) {
				for (Object x : setList.get(index2)) {
					setList.get(index1).add(x);
				}
				setList.remove(index2);
			}
			counts[a - 1]++;
			counts[b - 1]++;
		}

		for (int i = 0; i < k; i++) {
			int c = sc.nextInt();
			int d = sc.nextInt();
			int index1 = getSetIndex(setList, c);
			int index2 = getSetIndex(setList, d);
			if (setList.get(index1).contains(d)) {
				counts[c - 1]++;
			}
			if (setList.get(index2).contains(c)) {
				counts[d - 1]++;
			}
		}
		sc.close();

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n; i++) {
			int index = getSetIndex(setList, i + 1);
			int result = index == -1 ? 0 : setList.get(index).size() - 1 - counts[i];
			sb.append(i == 0 ? "" : " ");
			sb.append(result);
		}
		System.out.print(sb.toString());
	}

	private static int getSetIndex(ArrayList<HashSet> setList, int n) {
		for (int i = 0; i < setList.size(); i++) {
			if (setList.get(i).contains(n)) {
				return i;
			}
		}

		return -1;
	}
}