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# (.NET Core 3.1.201) (judge verdict: Accepted)

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

namespace ConsoleApp15
{
    class Class399
    {
        static void Main()
        {
            var nmk = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int N = nmk[0], M = nmk[1], K = nmk[2];
            var union = new Union_Find(N);
            var block = new List<int>[N];
            for (int i = 0; i < N; i++) block[i] = new List<int>();

            for(int i = 0; i < M; i++)
            {
                var ab = Console.ReadLine().Split().Select(_=>int.Parse(_)-1).ToArray();
                union.Unite(ab[0], ab[1]);
                block[ab[0]].Add(ab[1]);
                block[ab[1]].Add(ab[0]);
            }
            for (int i = 0; i < K; i++)
            {
                var cd = Console.ReadLine().Split().Select(_=>int.Parse(_)-1).ToArray();
                block[cd[0]].Add(cd[1]);
                block[cd[1]].Add(cd[0]);
            }

            int[] ans = new int[N];
            for(int i = 0; i < N; i++)
            {
                int num = union.getMem(i);
                foreach(int j in block[i])
                {
                    if (union.IsSameGroup(i, j))
                        num--;
                }
                ans[i] = num-1;
            }
            Console.WriteLine(string.Join(" ",ans));
        }
    }

    public class Union_Find
    {
        private int[] data;
        public Union_Find(int size)
        {
            data = new int[size];
            for (int i = 0; i < size; i++) data[i] = -1;
        }
        public bool Unite(int x, int y)
        {
            x = Root(x);
            y = Root(y);
            if (x != y)
            {
                if (data[y] < data[x])
                {
                    var tmp = y;
                    y = x;
                    x = tmp;
                }
                data[x] += data[y];
                data[y] = x;
            }
            return x != y;
        }
        public bool IsSameGroup(int x, int y) => Root(x) == Root(y);
        public int Root(int x) => data[x] < 0 ? x : data[x] = Root(data[x]);
        public int getMem(int x) => -data[Root(x)];
    }
}