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;
using System.Linq;

public class AtCoderD
{
    static void Main()
    {
        int n;
        int m;
        int k;
        {
            var line = Console.ReadLine();
            var words = line.Split(' ');
            n = int.Parse(words[0]);
            m = int.Parse(words[1]);
            k = int.Parse(words[2]);
        }

        UnionFind friendGroups = new UnionFind(n);
        int[] friendCount = new int[n + 1];
        for (int i = 0; i < m; i++)
        {
            var line = Console.ReadLine();
            var words = line.Split(' ');
            int a = int.Parse(words[0]);
            int b = int.Parse(words[1]);
            friendGroups.Unite(a, b);
            friendCount[a]++;
            friendCount[b]++;
        }

        int[] suggestions = new int[n];
        for (int i = 0; i < n; i++)
        {
            suggestions[i] = friendGroups.RootCount(i + 1) - friendCount[i + 1] - 1;
        }

        for (int i = 0; i < k; i++)
        {
            var line = Console.ReadLine();
            var words = line.Split(' ');
            int a = int.Parse(words[0]);
            int b = int.Parse(words[1]);
            if (friendGroups.SameRoot(a, b))
            {
                suggestions[a - 1]--;
                suggestions[b - 1]--;
            }
        }

        string answer = "";
        for (int i = 0; i < n; i++)
        {

            answer += " " + suggestions[i];
        }

        Console.WriteLine(answer.Trim());
    }

    public class UnionFind
    {
        private int[] parents;

        // 同じ根に属する枝の数を格納する
        private int[] rootCount;

        public int RootCount(int branch)
        {
            return rootCount[root(branch)];
        }

        public UnionFind(int count)
        {
            parents = new int[count + 1];
            rootCount = new int[count + 1];
            for (int i = 0; i < count + 1; i++)
            {
                parents[i] = i;
                rootCount[i] = 1;
            }
        }

        private int root(int N)
        {
            if (parents[N] == N) return N;
            return parents[N] = root(parents[N]);
        }

        //木を併合する際に、枝の数も合算する
        public void Unite(int X, int Y)
        {
            int rootX = root(X);
            int rootY = root(Y);
            if (rootX == rootY) return;
            rootCount[rootX] += rootCount[rootY];
            parents[rootY] = rootX;
        }

        public bool SameRoot(int X, int Y)
        {
            return root(X) == root(Y);
        }
    }
}