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: Accepted)

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();
    */

    @SuppressWarnings("unchecked")
    Set<Integer>[] g = new Set[N + 1];
    int[] fc = new int[N + 1];
    for (int i = 0; i < M; i++) {
      int a = sc.nextInt();
      int b = sc.nextInt();
      if (g[a] == null && g[b] == null) {
        Set<Integer> set = new HashSet<>();
        set.add(a);
        set.add(b);
        g[a] = set;
        g[b] = set;
      } else if (g[a] == null) {
        Set<Integer> set = g[b];
        set.add(a);
        g[a] = set;
      } else if (g[b] == null) {
        Set<Integer> set = g[a];
        set.add(b);
        g[b] = set;
      } else if (g[a] != g[b]) {
        Set<Integer> setL = g[a];
        Set<Integer> setS = g[b];
        if (setL.size() < setS.size()) {
          Set<Integer> temp = setL;
          setL = setS;
          setS = temp;
        }
        setL.addAll(setS);
        for (int c : setS) g[c] = setL;
      }
      fc[a]++;
      fc[b]++;
    }

    int[] bc = new int[N + 1];
    for (int i = 0; i < K; i++) {
      int c = sc.nextInt();
      int d = sc.nextInt();
      if (g[c] == g[d]) {
        bc[c]++;
        bc[d]++;
      }
    }
    sc.close();

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

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