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: Accepted)

import java.util.*;

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();

		StringBuilder ans = new StringBuilder();
		int[] edgenum = new int[N + 1];
		UnionFind uf = new UnionFind(N);

		// 友達関係を精査
		for (int i = 1; i <= M; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			uf.unite(a, b);
			edgenum[a]++;
			edgenum[b]++;
		}

		int[] x = new int[N + 1];
		// 各人の結果を計算
		for (int i = 1; i <= N; i++) {
			x[i] = uf.size(i) - edgenum[i] - 1;
		}
		// ブロック関係を精査
		for (int i = 1; i <= K; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			if (uf.isSame(a, b)) {
				x[a]--;
				x[b]--;
			}
		}
		for (int i = 1; i <= N; i++) {
			ans.append(x[i] + " ");
		}
		System.out.println(ans);

		sc.close();
	}

}

/**
 * UnionFind
 */
class UnionFind {

	int[] parent; // インデックス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保持
	int[] size; // 属しているグループのサイズを保持
	int[] rank; // parentと同様に、木の高さを保持

	public UnionFind(int n) {
		this.parent = new int[n + 1];
		this.size = new int[n + 1];
		this.rank = new int[n + 1];

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

	public void makeSet(int i) {
		parent[i] = i;
		size[i] = 1;
		rank[i] = 0;
	}

	public int size(int i) {
		return size[find(i)];
	}

	public void unite(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);

		if (xRoot == yRoot) {
			return;
		}
		// 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 {
			// yが属する木のランクがxのそれ以上の場合
			parent[xRoot] = yRoot;
			size[yRoot] += size[xRoot];
			if (rank[xRoot] == rank[yRoot]) {
				rank[xRoot]++;
			}
		}
	}

	// 引数の属する木のルートを返す
	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 isSame(int x, int y) {
		return find(x) == find(y);
	}
}