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 D_
{
    class Program
    {
        static void Main(string[] args)
        {
            var nmk = Console.ReadLine().Split(' ');
            var ninzuu = int.Parse(nmk[0]);
            var tomo = int.Parse(nmk[1]);
            var block = int.Parse(nmk[2]);
            var buffer = new int[ninzuu + 1];
            var uni = new UnionFind(ninzuu + 1);

            for(int i = 0; i < tomo; i++)
            {
                var line = Console.ReadLine().Split(' ');
                var x = int.Parse(line[0]);
                var y = int.Parse(line[1]);
                uni.Union(x, y);
                buffer[x]++;
                buffer[y]++;
            }
            for(int i = 0; i < block; i++)
            {
                var line = Console.ReadLine().Split(' ');
                var x = int.Parse(line[0]);
                var y = int.Parse(line[1]);
                if (uni.isSame(x, y))
                {
                    buffer[x]++;
                    buffer[y]++;
                }
            }
            for (int i = 1; i < ninzuu + 1; i++)
            {
                Console.Write(uni.Size(i) - buffer[i] - 1 + " ");
            }
            Console.ReadKey();
        }
    }

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

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

        public int Find(int x)
        {
            if (this.Parents[x] < 0) return x;

            this.Parents[x] = Find(this.Parents[x]);
            return this.Parents[x];
        }

        public int Size(int x)
        {
            return -this.Parents[this.Find(x)];
        }

        public bool isSame(int x,int y)
        {
            return this.Find(x) == this.Find(y);
        }

        public void Union(int x,int y)
        {
            x = this.Find(x);
            y = this.Find(y);
            if (x == y) return;
            if(Size(x) > Size(y))
            {
                var tmp = x;
                x = y;
                y = tmp;
            }
            this.Parents[y] += this.Parents[x];
            this.Parents[x] = y;
        }
    }
}