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: Time Limit Exceeded)

using System;
using System.Collections.Generic;

namespace ABC157d
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] s = Console.ReadLine().Split(' ');
            int N = int.Parse(s[0]);
            int M = int.Parse(s[1]);
            int K = int.Parse(s[2]);

            List<int> friend = new List<int>();
            var blockCount = new List<int>();
            for (int i = 0; i < N; i++)
            {
                friend.Add(0);
                blockCount.Add(0);
            }

            UnionFind tree = new UnionFind(N);

            for(int i = 0; i < M; i++)
            {
                s = Console.ReadLine().Split(' ');
                int a = int.Parse(s[0])-1;
                int b = int.Parse(s[1])-1;

                tree.Unite(a, b);
                friend[a] += 1;
                friend[b] += 1;
            }

            for (int i = 0; i < K; i++)
            {
                s = Console.ReadLine().Split(' ');
                int a = int.Parse(s[0]) - 1;
                int b = int.Parse(s[1]) - 1;

                if (tree.Same(a, b))
                {
                    blockCount[a] += 1;
                    blockCount[b] += 1;
                }
            }

            for (int i = 0; i < N; i++)
            {
                int sz = tree.Size(i);

                Console.Write((sz - 1 - friend[i] - blockCount[i]) + " ");
            }
         }

    }
    public class UnionFind
    {
        private List<int> par;
        private List<int> siz;

        public UnionFind(int N)
        {
            par = new List<int>();
            siz = new List<int>();
            for (int i = 0; i < N; i++)
            {
                par.Add(i);
                siz.Add(1);
            }
        }

        public int Root(int x)
        {
            if (par[x] == x) { return x; }

            return Root(par[x]);
        }

        public void Unite(int x, int y)
        {
            int rx = Root(x);
            int ry = Root(y);

            if (rx == ry) { return; }

            par[rx] = ry;
            siz[ry] += siz[rx];
        }
        public bool Same(int x, int y)
        {
            int rx = Root(x);
            int ry = Root(y);

            return rx == ry;
        }
        public int Size(int x)
        {
            return siz[Root(x)];
        }
    }
}