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.Diagnostics;
using System.Linq;

internal static class CR
{
    public static string GetText() { return Console.ReadLine(); }
    public static List<string> GetTexts() { return Console.ReadLine().Split(' ').ToList(); }
    public static long GetLong() { return long.Parse(Console.ReadLine()); }
    public static List<long> GetLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }
}

namespace AtCoder.D
{
    public class Program
    {
        public static long N { get; private set; }
        public static long M { get; private set; }
        public static long K { get; private set; }
        internal static List<AB> Friends { get; private set; }
        internal static List<CD> Blocks { get; private set; }

        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            var inputs = CR.GetLongs();
            N = inputs[0];
            M = inputs[1];
            K = inputs[2];
            Friends = new List<AB>();
            Blocks = new List<CD>();

            for (int i = 0; i < M; i++)
            {
                inputs = CR.GetLongs();
                Friends.Add(new AB
                {
                    A = inputs[0] - 1,
                    B = inputs[1] - 1,
                });
            }
            for (int i = 0; i < K; i++)
            {
                inputs = CR.GetLongs();
                Blocks.Add(new CD
                {
                    C = inputs[0] - 1,
                    D = inputs[1] - 1,
                });
            }

            var uf = new UnionFind((int)N);
            foreach (var item in Friends)
            {
                uf.Union((int)item.A, (int)item.B);
            }

            var result = new List<int>();
            for (int i = 0; i < N; i++)
            {
                var maybe = 0;
                for (int j = 0; j < N; j++)
                {
                    if (i == j) continue;

                    if (!Friends.Any(x => x.A == i && x.B == j || x.A == j && x.B == i)
                        && !Blocks.Any(x => x.C == i && x.D == j || x.C == j && x.D == i)
                        && uf.Same(i, j))
                    {
                        maybe++;
                    }
                }

                result.Add(maybe);
            }
            return string.Join(" ", result.Select(x => x));
        }
    }

    internal class CD
    {
        public long C { get; internal set; }
        public long D { get; internal set; }
    }

    internal class AB
    {
        public long A { get; internal set; }
        public long B { get; internal set; }
    }
    class UnionFind
    {
        // 親要素のインデックスを保持する
        // 親要素が存在しない(自身がルートである)とき、マイナスでグループの要素数を持つ
        public int[] Parents { get; set; }
        public UnionFind(int n)
        {
            this.Parents = new int[n];
            for (int i = 0; i < n; i++)
            {
                // 初期状態ではそれぞれが別のグループ(ルートは自分自身)
                // ルートなのでマイナスで要素数(1個)を保持する
                this.Parents[i] = -1;
            }
        }

        // 要素xのルート要素はどれか
        public int Find(int x)
        {
            // 親がマイナスの場合は自分自身がルート
            if (this.Parents[x] < 0) return x;
            // ルートが見つかるまで再帰的に探す
            // 見つかったルートにつなぎかえる
            this.Parents[x] = Find(this.Parents[x]);
            return this.Parents[x];
        }

        // 要素xの属するグループの要素数を取得する
        public int Size(int x)
        {
            // ルート要素を取得して、サイズを取得して返す
            return -this.Parents[this.Find(x)];
        }

        // 要素x, yが同じグループかどうか判定する
        public bool Same(int x, int y)
        {
            return this.Find(x) == this.Find(y);
        }

        // 要素x, yが属するグループを同じグループにまとめる
        public bool Union(int x, int y)
        {
            // x, y のルート
            x = this.Find(x);
            y = this.Find(y);
            // すでに同じグループの場合処理しない
            if (x == y) return false;

            // 要素数が少ないグループを多いほうに書き換えたい
            if (this.Size(x) < this.Size(y))
            {
                var tmp = x;
                x = y;
                y = tmp;
            }
            // まとめる先のグループの要素数を更新
            this.Parents[x] += this.Parents[y];
            // まとめられるグループのルートの親を書き換え
            this.Parents[y] = x;
            return true;
        }
    }
}