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;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;
using static System.Math;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;

public static class P
{
    public static void Main()
    {
        var nmk = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var n = nmk[0];
        var m = nmk[1];
        var k = nmk[2];
        UnionFind uf = new UnionFind(n);
        int[] member = new int[n];
        for (int i = 0; i < m; i++)
        {
            var ab = Console.ReadLine().Split().Select(x => int.Parse(x) - 1).ToArray();
            uf.TryUnite(ab[0], ab[1]);
            member[ab[0]]--;
            member[ab[1]]--;
        }
        for (int i = 0; i < k; i++)
        {
            var ab = Console.ReadLine().Split().Select(x => int.Parse(x) - 1).ToArray();
            if (uf.Find(ab[0]) != uf.Find(ab[1])) continue;
            member[ab[0]]--;
            member[ab[1]]--;
        }

        Console.WriteLine(string.Join(" ", Enumerable.Range(0, n).Select(x => uf.GetSize(x) + member[x] - 1)));
    }
}


class UnionFind
{
    public int Size { get; private set; }
    public int GroupCount { get; private set; }
    public IEnumerable<int> AllRepresents => Parent.Where((x, y) => x == y);
    int[] Sizes;
    int[] Parent;
    public UnionFind(int count)
    {
        Size = count;
        GroupCount = count;
        Parent = new int[count];
        Sizes = new int[count];
        for (int i = 0; i < count; i++) Sizes[Parent[i] = i] = 1;
    }
    public bool TryUnite(int x, int y)
    {
        int xp = Find(x);
        int yp = Find(y);
        if (yp == xp) return false;
        if (Sizes[xp] < Sizes[yp]) { var tmp = xp; xp = yp; yp = tmp; }
        GroupCount--;
        Parent[yp] = xp;
        Sizes[xp] += Sizes[yp];
        return true;
    }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public int GetSize(int x) => Sizes[Find(x)];
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public int Find(int x)
    {
        while (x != Parent[x]) x = (Parent[x] = Parent[Parent[x]]);
        return x;
    }
}