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;

class Program
{
    static void Main(string[] args)
    {
        var sol = new Solve();
        sol.Exec();
    }
}

class Solve
{
    public void Exec()
    {
        var l = Console.ReadLine().Split().Select(int.Parse).ToArray();
        int N = l[0], M = l[1], K = l[2];
        var uf = new UnionFind(N);
        var friend = new int[N];

        for (var i = 0; i < M; i++)
        {
            var ab = Console.ReadLine().Split().Select(int.Parse).ToArray();
            ab[0]--;
            ab[1]--;
            friend[ab[0]]++;
            friend[ab[1]]++;
            uf.Merge(ab[0], ab[1]);
        }

        var block = new List<List<int>>();
        for (var i = 0; i < N; i++) block.Add(new List<int>());
        for (var i = 0; i < K; i++)
        {
            var ab = Console.ReadLine().Split().Select(int.Parse).ToArray();
            ab[0]--;
            ab[1]--;
            block[ab[0]].Add(ab[1]);
            block[ab[1]].Add(ab[0]);
        }

        var ans = new List<int>();
        for (var i = 0; i < N; i++)
        {
            var t = uf.Size(i) - friend[i] - 1;
            foreach (var b in block[i])
            {
                if (uf.IsSame(i, b)) t--;
            }

            ans.Add(t);
        }

        Console.WriteLine(string.Join(" ", ans));
    }
}

class UnionFind
{
    List<int> par;

    public UnionFind(int n)
    {
        par = Enumerable.Repeat(-1, n).ToList();
    }

    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]) Swap(ref x, ref y);

        par[x] += par[y];
        par[y] = x;

        return true;
    }

    public int Size(int x)
    {
        return -par[Root(x)];
    }

    public void Swap(ref int a, ref int b)
    {
        var tmp = a;
        a = b;
        b = tmp;
    }
}