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.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static long Mod = (long)1e9 + 7;
        static void Main(string[] args)
        {
            //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            var NMK = Sarray();
            var N = NMK[0];
            var M = NMK[1];
            var K = NMK[2];
            // friend
            var friend = Enumerable.Range(0, (int)N).Select(_ => new List<long>()).ToArray();
            var funion = new UnionFind((int)N);
            for (var i = 0; i < M; ++i)
            {
                var ab = Sarray();
                friend[ab[0] - 1].Add(ab[1] - 1);
                friend[ab[1] - 1].Add(ab[0] - 1);
                funion.merge((int)ab[0] - 1, (int)ab[1] - 1);
            }
            // block
            var block = Enumerable.Range(0, (int)N).Select(_ => new List<long>()).ToArray();
            for (var i = 0; i < K; ++i)
            {
                var cd = Sarray();
                block[cd[0] - 1].Add(cd[1] - 1);
                block[cd[1] - 1].Add(cd[0] - 1);
            }
            var ans = new long[N];
            for (var i = 0; i < N; ++i)
            {
                var t = funion.size(i) - 1;//自分は除く
                t -= friend[i].Count;
                foreach (var b in block[i])
                    if (funion.issame(i, (int)b)) --t;
                ans[i] = t;
            }
            WriteLine(string.Join(" ", ans));
            //Out.Flush();
        }
    
        class UnionFind
        {
            int[] par;
            public UnionFind(int n)
            {
                init(n);
            }
            public void init(int n)
            {
                par = new int[n];
                for (var i = 0; i < n; ++i) { par[i] = -1; }
            }
            public int root(int x)
            {
                if (par[x] < 0) return x;
                else return par[x] = root(par[x]);
            }
            public bool issame(int x, int y)
            {
                return root(x) == root(y);
            }
            public bool merge(int x, int y)
            {
                x = root(x);
                y = root(y);
                if (x == y) return false;
                if (par[x] > par[y])
                {
                    var tmp = y;
                    y = x;
                    x = tmp;
                }
                par[x] += par[y];
                par[y] = x;
                return true;
            }
            public int size(int x)
            {
                return -par[root(x)];
            }
        }
    }
}