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.Linq;

namespace ABC157D
{
    class Program
    {
        static void Main(string[] args)
        {
            var NMK = ReadInt();
            int N = NMK[0];
            int M = NMK[1];
            int K = NMK[2];

            par = new int[N + 1];
            rank = new int[N + 1];
            init(N + 1);

            int[] friendCount = new int[N + 1];
            int[] brockCount = new int[N + 1];

            for (int i = 0; i < M; i++)
            {
                var ab = ReadInt();
                unite(ab[0], ab[1]);

                friendCount[ab[0]]++;
                friendCount[ab[1]]++;
            }

            for (int i = 0; i < K; i++)
            {
                var cd = ReadInt();
                if (!same(cd[0], cd[1])) continue;

                brockCount[cd[0]]++;
                brockCount[cd[1]]++;
            }

            int[] groupCount = new int[N + 1];
            for (int i = 1; i < N + 1; i++)
            {
                groupCount[find(i)]++;
            }

            for (int i = 1; i < N + 1; i++)
            {
                var cand = groupCount[find(i)] - friendCount[i] - brockCount[i] - 1;
                Console.WriteLine(cand);
            }
            Console.ReadKey();
        }

        //Union-find

        static int[] par;
        static int[] rank;

        static void init(int n)
        {
            for (int i = 0; i < n; i++)
            {
                par[i] = i;
                rank[i] = 0;
            }
        }

        static int find(int x)
        {
            if (par[x] == x)
            {
                return x;
            }
            else
            {
                return par[x] = find(par[x]);
            }
        }

        static void unite(int x, int y)
        {
            x = find(x);
            y = find(y);
            if (x == y) return;

            if (rank[x] < rank[y])
            {
                par[x] = y;
            }
            else
            {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }

        }
        static bool same(int x, int y)
        {
            return find(x) == find(y);
        }







        static int[] ReadInt()
        {
            int[] ret = Console.ReadLine().Split().Select(int.Parse).ToArray();
            return ret;
        }
        static long[] ReadLong()
        {
            long[] ret = Console.ReadLine().Split().Select(long.Parse).ToArray();
            return ret;
        }
    }
}