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;
using System.Collections.Generic;
using System.Collections.Specialized;
using System.Linq;
using System.Numerics;

namespace AtCoderWorkspace
{
    public class Solver
    {
        public void Solve()
        {
            var cin = new Scanner();
            var n = cin.nextInt();
            var m = cin.nextInt();
            var k = cin.nextInt();
            var friends = new int[n];
            var uni = new UnionFind(n);
            for (int i = 0; i < m; i++)
            {
                var a = cin.nextInt();
                var b = cin.nextInt();
                friends[a - 1]++;
                friends[b - 1]++;
                uni.Unite(a-1,b-1);
            }
            var block = new int[n];
            for (int i = 0; i < k; i++)
            {
                var a = cin.nextInt();
                var b = cin.nextInt();
                if (uni.Same(a-1, b-1))
                {
                    block[a - 1]++;
                    block[b - 1]++;
                }
            }
            Console.WriteLine(string.Join(" ", Enumerable.Range(0, n).Select(i => uni.GetSize(i) - friends[i] - block[i] - 1)));
        }
    }

    class UnionFind
    {
        public int Size { get; private set; }
        public int GroupCount { get; private set; }
        public IEnumerable<int> AllRepresents => Parent.Where((x, y) => x == y);
        int[] Sizes;
        int[] Parent;
        public UnionFind(int count)
        {
            Size = count;
            GroupCount = count;
            Parent = new int[count];
            Sizes = new int[count];
            for (int i = 0; i < count; i++) Sizes[Parent[i] = i] = 1;
        }
        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; }
            GroupCount--;
            Parent[yp] = xp;
            Sizes[xp] += Sizes[yp];
            return true;
        }
        public int GetSize(int x) => Sizes[Find(x)];
        public int Find(int x)
        {
            while (x != Parent[x]) x = (Parent[x] = Parent[Parent[x]]);
            return x;
        }
        public bool Same(int x, int y)
        {
            return Find(x) == Find(y);
        }
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }
        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public int[] DPArrayInt(int N, int ex)
        {
            int[] Array = new int[N + ex];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt();
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public long[] DPArrayLong(int N, int ex)
        {
            long[] Array = new long[N + ex];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong();
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }


        public double[] ArrayDouble(int N, double add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }

    class Program
    {
        static void Main(string[] s) => new Solver().Solve();
    }
}