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

class Program
{
    static void Main(string[] args) {
#if DEBUG
        MyDebugger.Test(process, "../TextFile1.txt");
        //MyDebugger.Test(process, "../TextFile2.txt");
        //MyDebugger.MakeTestCase("../TextFile2.txt");
#else
        object ret = process(new StreamReader(Console.OpenStandardInput()));
        Console.WriteLine(ret);
#endif
    }

    static void Write(object obj) {
#if DEBUG
        Debug.Write(obj.ToString());
#else
        Console.Write(obj.ToString());
#endif
    }

    //static long mod = 1000000007;
    static HashSet<int>[] friends;
    static HashSet<int>[] blocks;

    static object process(TextReader input) {
        var spl   = input.ReadLine().Split(' ');
        var n     = int.Parse(spl[0]);
        var m     = int.Parse(spl[1]);
        var k     = int.Parse(spl[2]);

        friends  = new HashSet<int>[n];
        blocks   = new HashSet<int>[n];

        for(int i=0; i<n; i++) {
            friends[i]  = new HashSet<int>();
            blocks[i]   = new HashSet<int>();
        }

        for(int i=0; i<m; i++) {
            spl   = input.ReadLine().Split(' ');
            var a = int.Parse(spl[0]) -1;
            var b = int.Parse(spl[1]) -1;
            friends[a].Add(b);
            friends[b].Add(a);
        }

        for(int i=0; i<k; i++) {
            spl   = input.ReadLine().Split(' ');
            var a = int.Parse(spl[0])-1;
            var b = int.Parse(spl[1])-1;
            blocks[a].Add(b);
            blocks[b].Add(a);
        }

        var graphs = new List<HashSet<int>>();
        var gno = new int[n];

        for(int i=0; i<n; i++) {
            gno[i] = -1;
        }

        for(int i=0; i<n; i++) {
            if(gno[i] < 0) {
                var gr = new HashSet<int>();
                NetWork(i, gr);
                graphs.Add(gr);
                foreach(var x in gr) {
                    gno[x] = graphs.Count - 1;
                }
            }
        }

        for(int i=0; i<n; i++) {
            var gr = graphs[gno[i]];

            int count = gr.Count;
            count -= friends[i].Count(x => gr.Contains(x));
            count -= blocks[i].Count(x => gr.Contains(x));
            count -= 1;

            Write(count.ToString());
            Write(' ');
        }

        return null;
    }

    private static void NetWork(int a, HashSet<int> box) {
        box.Add(a);

        foreach(var f in friends[a]) {
            if (!box.Contains(f)) {
                NetWork(f, box);
            }
        }
    }
}