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: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using static System.Math;
using Number = System.Int32;

namespace Program {
	public class Solver {
		Random rnd = new Random();
		public void Solve() {
			var n = ri;
			var m = ri;
			var k = ri;
			var s = new DisjointSet(n);
			var ans = new int[n];
			var G = Enumerate(n, x => new List<int>());
			for (int i = 0; i < m; i++) {
				var u = ri - 1;
				var v = ri - 1;
				s.Unite(u, v);
				ans[u]--;
				ans[v]--;
			}
			for (int i = 0; i < k; i++) {
				var u = ri - 1;
				var v = ri - 1;
				G[u].Add(v);
				G[v].Add(u);
			}
			for (int i = 0; i < n; i++) {
				ans[i] += s.Size(i) - 1;
				foreach (var x in G[i])
					if (s.Same(i, x)) ans[i]--;
			}
			Console.WriteLine(ans.AsJoinedString());
		}

		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n];
			for (int i = 0; i < a.Length; ++i) a[i] = f(i);
			return a;
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = true });
		var solver = new Program.Solver();
		//* 
		var t = new System.Threading.Thread(solver.Solve, 50000000);
		t.Start();
		t.Join();
		//*/
		// solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion


#region Disjoint Set
/// <summary>
/// いくつかの要素を素集合に分割して保持するコレクションです．
/// </summary>
public class DisjointSet {
	int[] par;
	byte[] rank;
	/// <summary>
	/// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
	/// </summary>
	/// <param name="N">要素数</param>
	/// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
	public DisjointSet(int N) {
		par = new int[N];
		rank = new byte[N];
		for (int i = 0; i < N; i++)
			par[i] = -1;
	}
	/// <summary>
	/// 指定した要素が属する集合の代表値を取得します．
	/// </summary>
	/// <param name="id">調べたい要素の 0-indexed での番号</param>
	/// <returns>指定した要素が属する集合の代表値</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int this[int id] {
		get {
			if ((par[id] < 0)) return id;
			return par[id] = this[par[id]];
		}
	}
	/// <summary>
	/// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>統合に成功したならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Unite(int x, int y) {
		x = this[x]; y = this[y];
		if (x == y) return false;
		if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
		par[x] += par[y];
		par[y] = x;
		if (rank[x] == rank[y])
			rank[x]++;
		return true;
	}
	/// <summary>
	/// 指定した要素が属する集合のサイズを取得します．
	/// </summary>
	/// <param name="x">指定する要素の 0-indexed での番号</param>
	/// <returns>集合のサイズ</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int Size(int x) { return -par[this[x]]; }

	/// <summary>
	/// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>同じならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Same(int x, int y) { return this[x] == this[y]; }

}
#endregion