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

namespace abc157_d
{
  class Program
  {
    static void Main(string[] args)
    {
      var N = CIn.ReadInt();
      var M = CIn.ReadInt();
      var K = CIn.ReadInt();

      var f = new int[N];
      var b = new HashSet<int>[N];

      for (int i = 0; i < N; i++)
      {
        b[i] = new HashSet<int>();
      }

      var uf = new UnionFind(N);

      for (int i = 0; i < M; i++)
      {
        var A = CIn.ReadInt() - 1;
        var B = CIn.ReadInt() - 1;

        f[A]++;
        f[B]++;

        uf.Unite(A, B);
      }

      for (int i = 0; i < K; i++)
      {
        var C = CIn.ReadInt() - 1;
        var D = CIn.ReadInt() - 1;

        b[C].Add(D);
        b[D].Add(C);
      }

      var ans = Enumerable.Range(0, N).Select(x => uf.GetSize(x) - 1 - f[x] - b[x].Count(y => uf.IsSame(x, y)));

      Console.WriteLine(String.Join(" ", ans));
    }
  }

  class UnionFind
  {
    private readonly int[] parent;
    private readonly int[] size;

    public UnionFind(int n)
    {
      parent = Enumerable.Range(0, n).ToArray();
      size = Enumerable.Repeat(1, n).ToArray();
    }

    public int GetRoot(int a)
    {
      if (parent[a] == a) return a;
      return parent[a] = GetRoot(parent[a]);
    }

    public void Unite(int a, int b)
    {
      var aRoot = GetRoot(a);
      var bRoot = GetRoot(b);

      if (aRoot == bRoot) return;

      if (size[a] >= size[b])
      {
        size[aRoot] += size[bRoot];
        parent[bRoot] = aRoot;
      }
      else
      {
        size[bRoot] += size[aRoot];
        parent[aRoot] = bRoot;
      }
    }

    public bool IsSame(int a, int b)
    {
      return GetRoot(a) == GetRoot(b);
    }

    public int GetSize(int a)
    {
      return size[GetRoot(a)];
    }
  }

  static class CIn
  {
    private static readonly Queue<string> fieldBuffer = new Queue<string>();

    public static string Read()
    {
      if (fieldBuffer.Count == 0)
      {
        foreach (var field in Console.ReadLine().Split())
        {
          fieldBuffer.Enqueue(field);
        }
      }
      return fieldBuffer.Dequeue();
    }

    public static int ReadInt() => int.Parse(Read());
    public static long ReadLong() => long.Parse(Read());
    public static double ReadDouble() => double.Parse(Read());

    public static string[] ReadStringArray(int n)
    {
      var array = new string[n];
      for (var i = 0; i < n; i++)
      {
        array[i] = Read();
      }
      return array;
    }

    public static int[] ReadIntArray(int n)
    {
      var array = new int[n];
      for (var i = 0; i < n; i++)
      {
        array[i] = ReadInt();
      }
      return array;
    }

    public static long[] ReadLongArray(int n)
    {
      var array = new long[n];
      for (var i = 0; i < n; i++)
      {
        array[i] = ReadLong();
      }
      return array;
    }

    public static double[] ReadDoubleArray(int n)
    {
      var array = new double[n];
      for (var i = 0; i < n; i++)
      {
        array[i] = ReadDouble();
      }
      return array;
    }

    public static char[,] ReadCharMatrix(int h, int w)
    {
      var array = new char[h, w];
      for (var i = 0; i < h; i++)
      {
        var l = Console.ReadLine().ToCharArray();
        for (var j = 0; j < w; j++)
        {
          array[i, j] = l[j];
        }
      }
      return array;
    }
  }
}