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.Windows;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Atcoder
{
    public class Program
    {
        static void Main(string[] args)
        {
            var r1 = Console.ReadLine().Split(' ');
            var N = r1[0].ToInt();
            var M = r1[1].ToInt();
            var K = r1[2].ToInt();

            var uf = new UnionFind(N);
            var edges_num = new int[N];
            var A = new int[M];
            var B = new int[M];
            
            for (int i = 0; i < M; i++)
            {
                var r2 = Console.ReadLine().Split(' ');
                var a = r2[0].ToInt();
                var b = r2[1].ToInt();
                A[i] = a;
                B[i] = b;
                A[i]--;
                B[i]--;
                edges_num[A[i]]++;
                edges_num[B[i]]++;
                uf.unite(A[i], B[i]);
            }
            var C = new int[K];
            var D = new int[K];

            var ans = new int[N];

            for (int i = 0; i < K; i++)
            {
                var r2 = Console.ReadLine().Split(' ');
                var c = r2[0].ToInt();
                var d = r2[1].ToInt();
                C[i] = c;
                D[i] = d;
                C[i]--;
                D[i]--;
                if (uf.isSame(C[i], D[i]))
                {
                    ans[C[i]]--;
                    ans[D[i]]--;
                }
            }

            for (int i = 0; i < N; i++)
            {
                ans[i] += uf.treeSize(i) - 1 - edges_num[i];
            }

            var sb = new StringBuilder();
            for (int i = 0; i < ans.Length; i++)
            {
                sb.Append(ans[i].ToString());
                if (i != (int)ans.Length - 1) sb.Append(" ");
            }
            Console.WriteLine(sb.ToString());
            Console.ReadLine();
        }
    }

    /* UnionFind：素集合系管理の構造体(union by size)
    isSame(x, y): x と y が同じ集合にいるか。 計算量はならし O(α(n))
    unite(x, y): x と y を同じ集合にする。計算量はならし O(α(n))
    treeSize(x): x を含む集合の要素数。
    */
    public class UnionFind
    {
        private int[] size;
        private int[] parents;
        public UnionFind() { }
        public UnionFind(int n)
        {
            // make n trees.
            size = new int[n];
            parents = new int[n];
            for (int i = 0; i < n; i++)
            {
                makeTree(i);
            }
        }
        public void makeTree(int x)
        {
            parents[x] = x;  // the parent of x is x
            size[x] = 1;
        }
        public bool isSame(int x, int y) { return findRoot(x) == findRoot(y); }
        public bool unite(int x, int y)
        {
            x = findRoot(x);
            y = findRoot(y);
            if (x == y) return false;
            if (size[x] > size[y])
            {
                parents[y] = x;
                size[x] += size[y];
            }
            else
            {
                parents[x] = y;
                size[y] += size[x];
            }
            return true;
        }
        public int findRoot(int x)
        {
            if (x != parents[x])
            {
                parents[x] = findRoot(parents[x]);
            }
            return parents[x];
        }
        public int treeSize(int x) { return size[findRoot(x)]; }
    }
}

public static class Extentions
{
    public static int ToInt(this object obj)
    {
        return int.Parse(obj.ToString());
    }

    public static decimal ToDecimal(this object obj)
    {
        return decimal.Parse(obj.ToString());
    }
}