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: Runtime Error)

using System;
using System.Collections;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;

namespace AtCoder
{
	public class ABC
	{
		private const int mod = 1000000007;  // 10^9+7

		static void Main(string[] args)
		{
			//System.Diagnostics.Stopwatch stopwatch = new System.Diagnostics.Stopwatch(); stopwatch.Start();
			Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });


			//string str = Console.ReadLine();
			//int n = int.Parse(Console.ReadLine());
			//long x = long.Parse(Console.ReadLine());
			int[] vs = Console.ReadLine().Split().Select(int.Parse).ToArray();
			//long[] vs = Console.ReadLine().Split().Select(long.Parse).ToArray();
			int n = vs[0];
			int m = vs[1];
			int k = vs[2];
			UnionFind tree = new UnionFind(n);
			int[,] matrix = new int[n, n];
			for (int i = 0; i < m; i++) {
				vs = Console.ReadLine().Split().Select(int.Parse).ToArray();
				int a = vs[0] - 1;
				int b = vs[1] - 1;
				tree.Unite(a, b);
				matrix[a, b] = 1;
				matrix[b, a] = 1;
			}
			for (int i = 0; i < k; i++) {
				vs = Console.ReadLine().Split().Select(int.Parse).ToArray();
				int c = vs[0] - 1;
				int d = vs[1] - 1;
				matrix[c, d] = -1;
				matrix[d, c] = -1;
			}
			for (int i = 0; i < n; i++) {
				int ans = 0;
				for (int j = 0; j < n; j++) {
					if (i != j && matrix[i, j] == 0 && tree.Same(i, j)) {
						ans++;
					}
				}
				Console.Write($"{ans} ");
			}
			Console.WriteLine();


			Console.Out.Flush();
			//System.Diagnostics.Debug.WriteLine(stopwatch.Elapsed);
		}
	}

	#region Union-Find木 UnionFind
	public class UnionFind
	{
		private readonly UnionFindNode[] _Parent;

		/// <summary>
		/// Union-Find木
		/// </summary>
		/// <param name="n">サイズ</param>
		public UnionFind(int n)
		{
			_Parent = new UnionFindNode[n];
			for (int i = 0; i < n; i++) {
				_Parent[i] = new UnionFindNode(i, 1);
			}
		}

		public int Find(int x)
		{
			if (_Parent[x].Parent == x) {
				return _Parent[x].Parent;
			}
			return _Parent[x].Parent = Find(_Parent[x].Parent);
		}

		public int Size(int x)
		{
			return _Parent[Find(x)].Size;
		}

		/// <summary>
		/// xとyの木を併合
		/// </summary>
		/// <param name="x"></param>
		/// <param name="y"></param>
		public void Unite(int x, int y)
		{
			int rx = Find(x);
			int ry = Find(y);
			if (rx != ry) {
				_Parent[ry].Parent = rx;
				_Parent[rx].Size += _Parent[ry].Size;
			}
		}

		/// <summary>
		/// x, yが同じ木に属している
		/// </summary>
		/// <param name="x"></param>
		/// <param name="y"></param>
		/// <returns></returns>
		public bool Same(int x, int y)
		{
			int rx = Find(x);
			int ry = Find(y);
			return rx == ry;
		}

		public override string ToString()
		{
			return string.Join(" ", _Parent.Select((p, i) => $"{i}({p.Parent},{p.Size})"));
		}
	}
	public class UnionFindNode
	{
		public int Parent;
		public int Size;
		public UnionFindNode(int parent, int size)
		{
			Parent = parent;
			Size = size;
		}
	}
	#endregion

}