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.*;

class Main{
  static class UnFd{
    int n;
    int forest;
    int[] parent;
    int[] size;
    UnFd(int n){
      forest=this.n=n;
      parent=new int[n];
      size=new int[n];
      for(int i=0;i<n;i++){
        parent[i]=i;
        size[i]=1;
      }
    }
    boolean isRoot(int i){
      return i==parent[i];
    }
    int rootOf(int i){
      if(isRoot(i)){
        return i;
      }
      return parent[i]=rootOf(parent[i]);
    }
    int sizeOf(int i){
      return size[rootOf(i)];
    }
    boolean same(int i,int j){
      return rootOf(i)==rootOf(j);
    }
    void merge(int i,int j){
      i=rootOf(i);
      j=rootOf(j);
      if(i==j){
        return;
      }
      size[i]=size[j]=size[i]+size[j];
      parent[Integer.max(i,j)]=Integer.min(i,j);
      forest--;
    }
    int[] parents(){
      return parent;
    }
    int[] sizes(){
      return size;
    }
    int forest(){
      return forest;
    }
    int length(){
      return n;
    }
  }
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int m=sc.nextInt();
    int k=sc.nextInt();
    UnFd uf=new UnFd(n);
    int[] minus=new int[n];
    for(int i=0;i<m;i++){
      int a=sc.nextInt()-1;
      int b=sc.nextInt()-1;
      uf.merge(a,b);
      minus[a]++;
      minus[b]++;
    }
    for(int i=0;i<k;i++){
      int c=sc.nextInt()-1;
      int d=sc.nextInt()-1;
      if(uf.same(c,d)){
        minus[c]++;
        minus[d]++;
      }
    }
    for(int i=0;i<n-1;i++){
      System.out.print((uf.sizeOf(i)-minus[i]-1)+" ");
    }
    System.out.println((uf.sizeOf(n-1)-minus[n-1]-1));
  }
}