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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
  
  list($n, $m, $k) = explode(" ", trim(fgets(STDIN)));

  $uf = new UnionFind($n);

  $friends = [];
  $block = [];

  for($i=1; $i<=$n; $i++){
    $friends[$i] = 0;
    $block[$i] = 0;
  }
  for($i=0; $i<$m; $i++){
    list($x, $y) = explode(" ", trim(fgets(STDIN)));
    $uf->unite($x, $y);
    $friends[$x]++;
    $friends[$y]++;
  }

  for($i=0; $i<$k; $i++){
    list($x, $y) = explode(" ", trim(fgets(STDIN)));
    if($uf->isSame($x, $y)){
      $block[$x]++;
      $block[$y]++;
    }
  }
  for($i=1; $i<=$n; $i++){
    echo $uf->size($i)-1-$friends[$i]-$block[$i];
    //echo $uf->size($i);
    if($i!=$n){
      echo " ";
    }
  }

  class UnionFind {
    public $d = [];
    function __construct($n){
      for($i=1;$i<=$n;$i++){
        $this->d[$i] = -1;
      }
    }
    function unite($x, $y){
      $rx = $this->root($x);
      $ry = $this->root($y);
      if($rx == $ry) return;
      //echo $rx." ".$ry." ";
      if($this->size($rx) > $this->size($ry)){
        $this->d[$rx]+= $this->d[$ry];
        $this->d[$ry] = $rx;
        //echo $this->d[$rx].PHP_EOL;
      }else{
        $this->d[$ry]+= $this->d[$rx];
        $this->d[$rx] = $ry;        
        //echo $this->d[$ry].PHP_EOL;
      }
    }
    function root($x){
      //echo $x.PHP_EOL;
      if($this->d[$x] < 0){
        return $x;
      }
      return $this->d[$x] = $this->root($this->d[$x]);
    }
    function size($x){
      return $this->d[$this->root($x)]*-1;
    }
    function isSame($x, $y){
      return $this->root($x) == $this->root($y);        
    }
  }