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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoders {
	class Program {
		static void Main(string[] args) {

          
          
			var inputArray = retreiveLongArray();
			var inputCount = 3;

			// 不正な入力は処理しない
			if (inputArray.Length != inputCount) {
				Console.WriteLine("invalid input");
				return;
			}
			long N = inputArray[0];
			long M = inputArray[1];
			long K = inputArray[2];

			//友達の情報を格納する
			var friendSet = new HashSet<int>[N];
			for (int i = 0; i < N; i++) {
				friendSet[i] = new HashSet<int>();
			}

			for (int i = 0; i < M; i++) {
				var friendPair = retreiveLongArray();
				var f1 = (int)friendPair[0] - 1;
				var f2 = (int)friendPair[1] - 1;

				friendSet[f1].Add(f2);
				friendSet[f2].Add(f1);
			}


			//ブロックしている人の情報を格納する
			var blockSet = new HashSet<long>[N];
			for (int i = 0; i < N; i++) {
				blockSet[i] = new HashSet<long>();
			}

			for (int i = 0; i < K; i++) {
				var blockPair = retreiveLongArray();
				var b1 = blockPair[0] - 1;
				var b2 = blockPair[1] - 1;

				blockSet[b1].Add(b2);
				blockSet[b2].Add(b1);
			}


			var nakayoshiGroups = new List<int[]>();
			var belongGroupDic = new Dictionary<int, int>();


			//友達の情報をもとに仲良しグループを作成する

			for (int i = 0; i < N; i++) {

				//自分がすでにどこかのグループに所属している場合は、処理しない
				if (belongGroupDic.ContainsKey(i)) {
					continue;
				}

				//グループを作成する
				var newGroup = Enumerable.Repeat(0, (int)N).ToArray();
				newGroup[i] = 1;
				nakayoshiGroups.Add(newGroup);
				var groupid = nakayoshiGroups.Count() - 1;
				belongGroupDic.Add(i, groupid);

				foreach (var friend in friendSet[i]) {
					addFriend(friend, groupid, newGroup, belongGroupDic, friendSet);
				}

	
			}


			var result = Enumerable.Repeat(0, (int)N).ToArray();//結果を格納する配列
			for (int i = 0; i < N; i++) {
				//どのグループに入っていない場合は0のまま
				if (!belongGroupDic.ContainsKey(i)) {
					continue;
				}

				//iはどのグループ?
				var groupId = belongGroupDic[i];
				var group = nakayoshiGroups[groupId];
				var kouho = group.Sum();//グループに所属している=1

				//ブロックしている人を取り除く
				foreach (var b in blockSet[i]) {
					var hantei = group[b];
					//ブロックしている人がグループにいる場合は候補を一人減らす
					if (hantei == 1) {
						kouho--;
					}
				}

				//友達の数と、自分自身を取り除く
				kouho = kouho - (friendSet[i].Count() + 1);

				result[i] = kouho;
			}
			//答えの出力
			var ans = string.Join(" ", result);
			Console.WriteLine(ans);

		}

		/// <summary>
		/// DFSで、友達グループを作る
		/// </summary>
		/// <param name="self"></param>
		/// <param name="groupId"></param>
		/// <param name="newGroup"></param>
		/// <param name="belongGroupDic"></param>
		/// <param name="friendSet"></param>
		private static void addFriend(int self, int groupId,int[] newGroup, Dictionary<int,int> belongGroupDic,HashSet<int>[] friendSet) {

			//既に処理済みの場合は、おしまい
			if (belongGroupDic.ContainsKey(self)){
				return;
			}

			belongGroupDic.Add(self, groupId);
			newGroup[self] = 1;

			foreach (var friend in friendSet[self]) {
				addFriend(friend, groupId, newGroup, belongGroupDic, friendSet);
			}
		}


		/// <summary>
		/// スペースで区切られた文字列配列の入力を受け取る
		/// </summary>
		/// <returns></returns>
		private static string[] retreiveStringArray() {
			var inputArray = Console.ReadLine().Split(' ');
			return inputArray;
		}

		/// <summary>
		/// スペースで区切られた整数配列の入力を受け取る
		/// </summary>
		/// <returns></returns>
		private static long[] retreiveLongArray() {
			var inputArray = Console.ReadLine().Split(' ');
			var longArray = inputArray.Select(s => long.Parse(s));
			return longArray.ToArray();
		}
	}
}