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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

//System.out.println();
public class Main implements Runnable { //クラス名はMain4

	public static void main(String[] args) {
		Thread.setDefaultUncaughtExceptionHandler((t, e) -> System.exit(1));
		new Thread(null, new Main(), "", 16 * 1024 * 1024).start(); //16MBスタックを確保して実行
	}

	public void run() {
		FastScannerMain4 sc = new FastScannerMain4(System.in);

		try {
			int N = sc.nextInt();
			int M = sc.nextInt();
			int K = sc.nextInt();
			int[][] fri = new int[N + 1][N + 1];
			int[][] block = new int[N + 1][N + 1];

			//UnionFindでGroupを作成
			UnionFindTree unionfindtreeFriends = new UnionFindTree(N);

			for (int i = 0; i < M; i++) {
				int a = sc.nextInt();
				int b = sc.nextInt();
				fri[a - 1][b - 1] = 1;
				fri[b - 1][a - 1] = 1;
				if (!unionfindtreeFriends.same(a - 1, b - 1)) {
					unionfindtreeFriends.unite(a - 1, b - 1);
				}
			}

			for (int i = 0; i < K; i++) {
				int c = sc.nextInt();
				int d = sc.nextInt();
				block[c - 1][d - 1] = 1;
				block[d - 1][c - 1] = 1;
			}

			int anseach = 0;
			String ans = "";
			for (int i = 0; i < N; i++) {
				//友達候補から自分自身を引くので-1とする。
				anseach = unionfindtreeFriends.size(i) - 1;

				for (int j = 0; j < N; j++) {
					//Blockかつ、同じGroupの場合、-1
					if (block[i][j] > 0 && unionfindtreeFriends.same(i, j)) {
						anseach--;
					}
					//Friendかつ、同じGroupの場合、-1
					if (fri[i][j] > 0 && unionfindtreeFriends.same(i, j)) {
						anseach--;
					}
				}

				if (i == N - 1) {
					ans += anseach;
				} else {
					ans += anseach + " ";
				}
			}

			System.out.println(ans);

		} catch (Exception e) {
			e.printStackTrace();
		}

	}
}

class UnionFindTree {

	int[] parent; //インデックスにとノードを対応させ、そのルートノードのインデックスを格納
	int[] rank; //parentと同様に、木の高さを格納
	int[] size; //木の要素の数を格納。最新はparentのindexに
	int group;

	public UnionFindTree(int n) {
		this.parent = new int[n];
		this.rank = new int[n];
		this.size = new int[n];
		group = n;

		for (int i = 0; i < n; i++) {
			makeSet(i);
		}
	}

	public void makeSet(int i) {
		parent[i] = i;
		rank[i] = 0; //集合の高さ
		size[i] = 1;
	}

	public void unite(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);
		if (xRoot == yRoot)
			return;
		group--;
		//xが属する木の方が大きい場合
		if (rank[xRoot] > rank[yRoot]) {
			parent[yRoot] = xRoot; //yの親をxに更新
			size[xRoot] += size[yRoot];

		} else if (rank[xRoot] < rank[yRoot]) {
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
		} else {
			parent[yRoot] = xRoot;
			rank[xRoot]++; //同じ高さの木がルートの子として着くから大きさ++;
			size[xRoot] += size[yRoot];
		}
	}

	//引数の属する木のルートのidを返す
	public int find(int i) {
		if (i != parent[i]) {
			parent[i] = find(parent[i]);
		}
		return parent[i];
	}

	//同じ木に属しているか
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}

	// via Oland https://atcoder.jp/contests/abc120/submissions/4452920
	public int size(int i) {
		return size[find(i)];
	}
}

//高速なScanner
class FastScannerMain4 {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScannerMain4(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}