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: Wrong Answer)

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

class Program
{
    static void Main(string[] args)
    {
        int[] arrMoto = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
        int n = arrMoto[0];
        int m = arrMoto[1];
        int k = arrMoto[2];
        int[] ans = new int[n];
        Person[] arrp = new Person[n];
        for (int i = 1; i <= n; i++) arrp[i - 1] = new Person(i);
        for(int i = 1; i <= m; i++)
        {
            int[] arrMotoTwo = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
            int a = arrMotoTwo[0];
            int b = arrMotoTwo[1];
            arrp[a - 1].friend.Add(arrp[b - 1]);
            arrp[b - 1].friend.Add(arrp[a - 1]);
        }
        for (int i = 1; i <= k; i++)
        {
            int[] arrMotoTwo = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
            int c = arrMotoTwo[0];
            int d = arrMotoTwo[1];
            arrp[c - 1].block.Add(arrp[d - 1]);
            arrp[d - 1].block.Add(arrp[c - 1]);
        }
        foreach(Person person in arrp)
        {
            foreach(Person other in person.other(arrp,person))
            {
                if (person.no > other.no) break;
                if(other.no <= person.no + person.friendLen(arrp))
                {
                    ans[person.no - 1]++;
                    ans[other.no - 1]++;
                } else
                {
                    break;
                }
            }
        }
        foreach (int ele in ans)
            Console.Write("{0} ", ele);

    }

}

class Person
{
    public int no;
    public Person(int no)
    {
        this.no = no;
    }
    public List<Person> friend = new List<Person>();
    public List<Person> block = new List<Person>();
    public List<Person> other(Person[] arrp,Person own)
    {
        var me = arrp.ToList();
        me.Add(own);
        var fb = friend.Concat(block);
        return me.Concat(fb).GroupBy(v => v).Where(v => v.Count() == 1).Select(v => v.Key).OrderBy(v => v.no).ToList();
    }

    public int friendLen(Person[] arrp)
    {
        int len = 0;
        int i = this.no - 1;
        while (i < arrp.Length - 1)
        {
            if (arrp[i].friend.Contains(arrp[i + 1])) len++;
            else break;
            i++;
        }
        return len;
    }


}