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.Collections;
using System.Linq;
namespace Atcoder_0404
{
    class Program
    {
        static void Main(string[] args)
        {
            var NMK = GetInput_long();
            long _humanCount = NMK[0];
            long _fraiendCount = NMK[1];
            long _blockCount = NMK[2];

            List<long>[] _friendList = new List<long>[_humanCount];
            List<long>[] _blockList = new List<long>[_humanCount];
            HashSet<long>[] _newFraiendList = new HashSet<long>[_humanCount];
            HashSet<long>[] _seenList = new HashSet<long>[_humanCount];
            for (int i =0; i < _friendList.Length; i++)
            {
                _friendList[i] = new List<long>();
                _blockList[i] = new List<long>();
                _newFraiendList[i] = new HashSet<long>();
                _seenList[i] = new HashSet<long>();
            }

            for (int i = 0; i < _fraiendCount; i++)
            {
                var input = GetInput_long();
                var a = input[0];
                var b = input[1];
                _friendList[a-1].Add(b-1);
                _friendList[b-1].Add(a-1);
            }
            for (int i = 0; i < _blockCount; i++)
            {
                var input = GetInput_long();
                var a = input[0];
                var b = input[1];
                _blockList[a-1].Add(b-1);
                _blockList[b-1].Add(a-1);
            }
            //ここから実装
            for (int i = 0; i < _humanCount; i++)
            {
                Stack<long> stack = new Stack<long>();
                var start = i;
                stack.Push(start);
                while (stack.Count > 0)
                {
                    var target = stack.Pop();
                    _seenList[i].Add(target);
                    foreach(var connect in _friendList[target])
                    {
                        if (!_seenList[i].Contains(connect))
                        {
                            stack.Push(connect);
                        }
                    }
                    if(!_friendList[i].Contains(target)
                        && !_blockList[i].Contains(target))
                    {
                        _newFraiendList[i].Add(target);
                    }
                }
            }
            for (int i = 0; i < _humanCount; i++)
            {
                Console.Write(_newFraiendList[i].Count-1);
                if (i < _humanCount - 1) Console.Write(" ");
                else Console.Write("\n");
            }
        }
        #region input
        

        static string[] GetInput_string()
        {
            var input = Console.ReadLine().Split(' ');
            return input;
        }

        static int[] GetInput_int()
        {
            var input = GetInput_string();
            var result = new int[input.Length];
            for (int i = 0; i < result.Length; i++)
            {
                result[i] = int.Parse(input[i]);
            }
            return result;
        }
        static long[] GetInput_long()
        {
            var input = GetInput_string();
            var result = new long[input.Length];
            for (int i = 0; i < result.Length; i++)
            {
                result[i] = long.Parse(input[i]);
            }
            return result;
        }
        #endregion
        #region sonota
        static void Swap_array<T>(T[] target, int a, int b)
        {
            T temp = target[a];
            target[a] = target[b];
            target[b] = temp;
        }
        static void Swap_list<T>(List<T> target, int a, int b)
        {
            T temp = target[a];
            target[a] = target[b];
            target[b] = temp;
        }
        /// <summary>
        /// 最大公約数
        /// </summary>
        public static int GCD(int a, int b)
        {

            /* 自然数 a > b を確認・入替 */
            if (a < b)
            {
                int tmp = a;
                a = b;
                b = tmp;
            }

            /* ユークリッドの互除法 */
            int r = a % b;
            while (r != 0)
            {
                a = b;
                b = r;
                r = a % b;
            }
            return b;
        }

        /// <summary>
        /// 最小公倍数
        /// </summary>
        public static int LCM(int a, int b)
        {
            int x = a * b;
            b = GCD(a, b);
            return x / b;

        }
        #endregion

        static long mod = 1000000000 + 7;
    }
}