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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import static java.lang.System.*;
import java.io.*;
import java.math.*;

public class Main {
		static List<Integer> passed = new ArrayList<Integer>();
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		int[] ff = new int[n+1];
		Account[] accList = new Account[n + 1];
		for(int i = 1; i <= n; i++) {
			accList[i] = new Account();
		}
		for(int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			accList[a].fList.add(b);
			accList[b].fList.add(a);
		}
		for(int i = 0; i < k; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			accList[a].bList.add(b);
			accList[b].bList.add(a);
		}
		/*
		for(int i = 1; i <= n; i++) {
			passed.clear();
			DFS(i, i,accList);
			ff[i] += passed.size() - 1;	
			out.println("ff[" + i + "]:" + ff[i]) ;
		}
		*/
		/*
		for(int i =1; i <=n; i++) {
			Account a = accList[i];
			out.print(i + "f:");
			for(int j = 0; j < a.fList.size(); j++) {
				out.print(a.fList.get(j) + " ");
			}
			out.print(" b:");
			for(int j = 0; j < a.bList.size(); j++) {
				out.print(a.bList.get(j) + " ");
			}
			out.println();
		}*/
		
		for( int i = 1; i <= n; i++) {
			long cnt = 0;
			cnt += ff[i] - accList[i].fList.size() ;
			out.print(cnt + " ");
		}

		out.close();
	}
	static void DFS(int node,  int root, Account[] accList) {
		if(!accList[root].bList.contains(node))
		passed.add(node);
		for(int i = 0; i < accList[node].fList.size(); i++) {
			int num = accList[node].fList.get(i);
			if (passed.contains(num))
				continue;
			DFS(num, root, accList);
		}
	}
	
}
class Account {

	List<Integer> fList = new ArrayList<>();
	List<Integer> bList = new ArrayList<>();

}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}