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: Time Limit Exceeded)

import java.util.*;
import java.math.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int K = sc.nextInt();
    int[] A = new int[M];
    int[] B = new int[M];
    for (int i = 0; i < M; i++) {
      A[i] = sc.nextInt();
      B[i] = sc.nextInt();
    }
    int[] C = new int[K];
    int[] D = new int[K];
    for (int i = 0; i < K; i++) {
      C[i] = sc.nextInt();
      D[i] = sc.nextInt();
    }
    sc.close();


    int[] fc = new int[N + 1];
    Map<Integer, Set<Integer>> g = new HashMap<>();
    for (int i = 0; i < M; i++) {
      Integer a = Integer.valueOf(A[i]);
      Integer b = Integer.valueOf(B[i]);
      if (!g.containsKey(a) && !g.containsKey(b)) {
        Set<Integer> set = new HashSet<>();
        set.add(a);
        set.add(b);
        g.put(a, set);
        g.put(b, set);
      } else if (!g.containsKey(a)) {
        Set<Integer> set = g.get(b);
        set.add(a);
        g.put(a, set);
      } else if (!g.containsKey(b)) {
        Set<Integer> set = g.get(a);
        set.add(b);
        g.put(b, set);
      } else {
        Set<Integer> set = new HashSet<>();
        set.addAll(g.get(a));
        set.addAll(g.get(b));
        for (Integer c : set)
          g.put(c, set);
      }
      fc[a]++;
      fc[b]++;
    }

    int[] bc = new int[N + 1];
    for (int i = 0; i < K; i++) {
      Integer c = Integer.valueOf(C[i]);
      Integer d = Integer.valueOf(D[i]);
      if (g.get(c) == g.get(d)) {
        bc[c]++;
        bc[d]++;
      }
    }

    int[] ans = new int[N + 1];
    for (int i = 1; i < N + 1; i++)
      if (g.get(i) != null)
        ans[i] = g.get(i).size() - 1 - fc[i] - bc[i];

    for (int i = 1; i < N + 1; i++)
      System.out.print(ans[i] + " ");
    System.out.println();
  }
}