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 static System.Console;
using static System.Math;

namespace D_ABC157
{
    class MainClass
    {

        static int[] s;

        public static void Main(string[] args)
        {
            s = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
            int n = s[0];
            int m = s[1];
            int k = s[2];
            var ab = new List<int[]>();
            var cd = new List<int[]>();

            var uf = new UnionFind(n);

            int[] cntlst = new int[n];

            for (int i = 0; i < m; i++)
            {
                int[] tmp = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
                // ab.Add(tmp);
                uf.Unite(tmp[0] - 1, tmp[1] - 1);
                cntlst[tmp[0] - 1]++;
                cntlst[tmp[1] - 1]++;


            }
            for (int i = 0; i < k; i++)
            {
                int[] tmp = Array.ConvertAll(ReadLine().Split().ToArray(), int.Parse);
                // cd.Add(tmp);
                if (uf.Find(tmp[0] - 1) == uf.Find(tmp[1] - 1))
                {
                    cntlst[tmp[0] - 1]++;
                    cntlst[tmp[1] - 1]++;

                }
            }

            for (int i = 0; i < n; i++)
            {
                Write(uf.GetSize(i) - cntlst[i] - 1);
                if (i != n - 1)
                {
                    Write(" ");
                }
                else
                {
                    WriteLine();
                }





            }







        }
    }


    class UnionFind
    {
        public int Size { get; private set; }
        List<int> parent;
        List<int> sizes;
        public UnionFind(int count)
        {
            Size = 0;
            parent = new List<int>(count);
            sizes = new List<int>(count);
            ExtendSize(count);
        }
        public bool Unite(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;
            }
            parent[yp] = xp;
            sizes[xp] += sizes[yp];
            return true;
        }
        public IEnumerable<int> AllParents => parent.Where(x => x == parent[x]);
        public int GetSize(int x) => sizes[Find(x)];
        public int Find(int x) => parent[x] == parent[parent[x]] ? parent[x] : parent[x] = Find(parent[x]);
        public void ExtendSize(int treeSize)
        {
            if (treeSize <= Size) return;
            parent.Capacity = treeSize;
            sizes.Capacity = treeSize;
            while (Size < treeSize)
            {
                parent.Add(Size);
                sizes.Add(1);
                Size++;
            }
        }
    }
}