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;
using System.Text;
using System.Threading.Tasks;

namespace Atcoder
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] nmk = Console.ReadLine().Split(' ');
            int n = int.Parse(nmk[0]);
            int m = int.Parse(nmk[1]);
            int k = int.Parse(nmk[2]);
            int[] a = new int[m];
            int[] b = new int[m];
            int[] yujins = new int[n];
            UnionFind uni = new UnionFind(n);


            int[] blocking = new int[n];
            for (int i = 0; i < m; i++)
            {
                string[] ab = Console.ReadLine().Split(' ');
                a[i] = int.Parse(ab[0])-1;
                b[i] = int.Parse(ab[1])-1;
                yujins[a[i]]++;
                yujins[b[i]]++;
                uni.Unite(a[i], b[i]);
            }
            int[] c = new int[k];
            int[] d = new int[k];
            for (int i = 0; i < k; i++)
            {
                string[] ab = Console.ReadLine().Split(' ');
                c[i] = int.Parse(ab[0])-1;
                d[i] = int.Parse(ab[1])-1;

                if(uni.IsSameGroup(c[i],d[i]))
                {
                    blocking[c[i]]++;
                    blocking[d[i]]++;
                }
            }
            for (int i = 0; i < n; i++)
            {
                int kore = uni.Root(i);
                int kouhos = uni.data[uni.Root(i)];
                Console.Write(-kouhos - yujins[i] - blocking[i]-1);
                if (i != n - 1) { Console.Write(" "); }
            }
            Console.WriteLine();

            
        }
    }

    public class UnionFind
    {
        public int[] data;

        public UnionFind(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)
        {
            return Root(x) == Root(y);
        }

        public int Root(int x)
        {
            return data[x] < 0 ? x : data[x] = Root(data[x]);
        }
    }
}