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.IO;

namespace AtCoder
{
	class Program
	{
		public static void Main()
		{
#if TextInput
			using(var reader = new StreamReader(Debug.SampleTextPath)) 
			{
				Console.SetIn(reader);
#endif
				var intArray = IO.ReadIntArray();
				var answerNums = new int[intArray[0] + 1];
				var unionFind = new UnionFind(intArray[0]);
				for(var iIndex = 0 ; iIndex < intArray[1] ; ++ iIndex)
				{
					var friendIntArray = IO.ReadIntArray();
					unionFind.Union(friendIntArray[0] , friendIntArray[1]);
					-- answerNums[friendIntArray[0]];
					-- answerNums[friendIntArray[1]];
				}
				unionFind.Normalize();

				for(var iIndex = 0 ; iIndex < intArray[2] ; ++ iIndex)
				{
					var blockIntArray = IO.ReadIntArray();
					if(unionFind.Parents[blockIntArray[0]] == unionFind.Parents[blockIntArray[1]])
					{
						-- answerNums[blockIntArray[0]];
						-- answerNums[blockIntArray[1]];
					}
				}

				for(var iIndex = 1 ; iIndex <= intArray[0] ; ++ iIndex)
				{
					answerNums[iIndex] += (unionFind.RootNums[unionFind.Parents[iIndex]] - 1);
				}

				IO.Write(answerNums.Skip(1).ToArray());
#if TextInput
			}
#endif
		}

		private class UnionFind
		{
			public int[] Parents;
			public int[] RootNums;
			public UnionFind(int iRootNum)
			{
				Parents = Enumerable.Range(0 , iRootNum + 1).ToArray();
				RootNums = new int[iRootNum + 1];
			}
			public void Normalize()
			{
				for(var iIndex = 1 ; iIndex < Parents.Length ; ++ iIndex)
				{
					var rootIndex = _GetRoot(iIndex);
					++ RootNums[rootIndex];
					Parents[iIndex] = rootIndex;
				}
			}
			public void Union(int iNumber0 , int iNumber1)
			{
				var rootNumber0 = _GetRoot(iNumber0);
				var rootNumber1 = _GetRoot(iNumber1);
				if(rootNumber0 < rootNumber1)
				{
					Parents[rootNumber1] = rootNumber0;
				}
				else
				{
					Parents[rootNumber0] = rootNumber1;
				}
			}
			public int _GetRoot(int iNumber)
			{
				return (Parents[iNumber] == iNumber) ? iNumber : _GetRoot(Parents[iNumber]);
			}
		}
	}

	public static class IO
	{
		public static string   Read()								 { return Console.ReadLine(); }
		public static string[] ReadArray(char iSplitter = ' ')		 { return Read().Split(iSplitter); }
		public static int	   ReadInt()							 { return int.Parse(Read()); }
		public static int[]	   ReadIntArray(char iSplitter = ' ')	 { return ReadArray().Select(iText => int.Parse(iText)).ToArray(); }
		public static long	   ReadLong()							 { return int.Parse(Read()); }
		public static long[]   ReadLongArray(char iSplitter = ' ')	 { return ReadArray().Select(iText => long.Parse(iText)).ToArray(); }
		public static double   ReadDouble()							 { return double.Parse(Read()); }
		public static double[] ReadDoubleArray(char iSplitter = ' ') { return ReadArray().Select(iText => double.Parse(iText)).ToArray(); }
		public static char[]   ReadCharArray()						 { return Read().ToCharArray(); }

		public static void Write(string iValue)									   { Console.WriteLine(iValue); }
		public static void Write(int iValue)									   { Console.WriteLine(iValue); }
		public static void Write(IEnumerable<int> iValue , string iJoinWord = " ") { Console.WriteLine(string.Join(iJoinWord , iValue)); }
		public static void Write(long iValue)									   { Console.WriteLine(iValue); }
		public static void Write(double iValue)									   { Console.WriteLine(iValue); }
		public static void Write(decimal iValue)								   { Console.WriteLine(iValue); }
		public static void Write(bool iValue)									   { Console.WriteLine(iValue ? "Yes" : "No"); }
	}

	public static class Util
	{
		public static long Gcd(long i0 , long i1) { return (i0 > i1) ? _Gcd(i0 , i1) : _Gcd(i1 , i0); }
		public static long Lcm(long i0 , long i1) { return (i0 * i1) / Gcd(i0 , i1); }
		public static long ModPow(long iBaseNumber , long iPow , long iModNumber)
		{
			var num = 1L;
			while(iPow != 0)
			{
				if((iPow & 0x1) != 0)
				{
					num = (num * iBaseNumber) % iModNumber;
				}
				iBaseNumber = (iBaseNumber * iBaseNumber) % iModNumber;
				iPow >>= 1;
			}
			return num;
		}
		public static long CombinationNum(int iLeftNumber , int iRightNumber)
		{
			if(iRightNumber == 0) { return 1; }
			if(iLeftNumber == 0)  { return 0; }
			return iLeftNumber * CombinationNum(iLeftNumber - 1 , iRightNumber - 1) / iRightNumber;
		}

		private static long _Gcd(long i0 , long i1)	{ return (i1 == 0) ? i0 : _Gcd(i1 , i0 % i1); }
	}
}