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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
  static int[] par;
  static int[] rank;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    int k = sc.nextInt();
    par = new int[n + 1];
    rank = new int[n + 1];

    int[] f = new int[n + 1];
    init(n + 1);
    for(int i = 0; i < m; i++) {
      int a = sc.nextInt();
      int b = sc.nextInt();
      f[a]++;
      f[b]++;
      unite(a, b);
    }

    int[] ans = new int[n + 1];

    int[] num = new int[n + 1];

    for(int i = 1; i <= n; i++) {
      num[find(i)]++;
    }
    
    for(int i = 1; i <= n; i++) {
      ans[i] = num[find(i)] - f[i] - 1;
    }

    for(int i = 0; i < k; i++) {
      int c = sc.nextInt();
      int d = sc.nextInt();
      if(same(c, d)) {
        ans[c]--;
        ans[d]--;
      }
    }

    for(int i = 1; i < n + 1; i++) {
      System.out.print(ans[i]);
      System.out.print(" ");
    }
  }
  static void init(int m) {
      for(int i = 0; i < m; i++) {
        par[i] = i;
        rank[i] = 1;
      }
    }
  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) {
        if(rank[x] < rank[y]) {
          par[x] = y;
          rank[y] += rank[x];
        } else {
          if(rank[x] == rank[y]) {
            if(x > y) {
              par[x] = y;
              rank[y] += rank[x];
            } else {
              par[y] = x;
              rank[x] += rank[y];
            }
          } else {
            par[y] = x;
            rank[x] += rank[y];
          }
        }
      }
    }
  static boolean same(int x, int y) {
      return find(x) == find(y);
    }
}