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;

class Program
{
    static void Main(string[] args)
    {
        var line = Console.ReadLine().Split(' ');
        var personNumber = int.Parse(line[0]);
        var friendNumber = int.Parse(line[1]);
        var blockNumber = int.Parse(line[2]);

        var unionFind = new UnionFind(personNumber);

        var list = new List<int[]>();

        var result = new int[personNumber];

        for (var i = 0; i < friendNumber; i++)
        {
            var l = Console.ReadLine().Split(' ');
            unionFind.UniteMembers(int.Parse(l[0]) - 1, int.Parse(l[1]) - 1);

            list.Add(new int[2] { int.Parse(l[0]) - 1, int.Parse(l[1]) - 1 });

        }

        for (var i = 0; i < blockNumber; i++)
        {
            var l = Console.ReadLine().Split(' ');
            list.Add(new int[2] { int.Parse(l[0]) - 1, int.Parse(l[1]) - 1 });
        }


        for (var i = 0; i < personNumber; i++)
        {
            result[i] = unionFind.GetMenberSize(i) - 1;
        }


        foreach (var item in list)
        {
            if (unionFind.CheckSameGroup(item[0], item[1]))
            {
                result[item[0]] -= 1;
                result[item[1]] -= 1;
            }        }
        Console.WriteLine(string.Join(" ", result));
    }
}

class UnionFind
{
    public int[] ParentId { get; set; }

    public UnionFind(int nodeNumber)
    {
        this.ParentId = new int[nodeNumber];
        for (var i = 0; i < nodeNumber; i++)
        {
            this.ParentId[i] = -1;
        }
    }

    public int FindParentId(int id)
    {
        if (this.ParentId[id] < 0)
        {
            return id;
        }

        this.ParentId[id] = this.FindParentId(this.ParentId[id]);
        return this.ParentId[id];
    }

    public int GetMenberSize(int id)
    {
        var parentId = this.FindParentId(id);
        return -this.ParentId[parentId];
    }

    public bool CheckSameGroup(int x, int y)
    {
        return this.FindParentId(x) == this.FindParentId(y);
    }

    public void UniteMembers(int x, int y)
    {
        var xParentId = this.FindParentId(x);
        var yParentId = this.FindParentId(y);

        if (xParentId == yParentId)
        {
            return;
        }


        if (this.ParentId[xParentId] < this.ParentId[xParentId])
        {
            var tmp = xParentId;
            xParentId = yParentId;
            yParentId = tmp;
        }

        this.ParentId[xParentId] += this.ParentId[yParentId];
        this.ParentId[yParentId] = xParentId;
    }

}