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: Wrong Answer)

<?php

list($n, $m, $k) = explode(' ', trim(fgets(STDIN)));

$relation = [];
for ($i = 1; $i <= $m; ++$i) {
    list($a, $b) = explode(' ', trim(fgets(STDIN)));
    $relation[$a][$b] = 'friend';
    $relation[$b][$a] = 'friend';
}

for ($i = 1; $i <= $k; ++$i) {
    list($c, $d) = explode(' ', trim(fgets(STDIN)));
    $relation[$c][$d] = 'block';
    $relation[$d][$c] = 'block';
}

$candidates = [];
for ($i = 1; $i <= $n; ++$i) {
    for ($j = $i + 1; $j <= $n; ++$j) {
        $num = 0;

        if (isset($relation[$i][$j])) {
            continue;
        }

        if (findFriends($friends, $i, $j)) {
            ++$num;
        }

        $candidates[$i] = $num;
    }
}

echo implode(' ', $candidates);

function findFriends($relation, $a, $b)
{
    if (isset($relation[$a][$b]) && $relation[$a][$b] === 'friend') {
        return true;
    }

    foreach ($relation[$a] as $c => $a_c_relation) {
        if ($a_c_relation === 'friend'){
            return findFriends($relation, $c, $b);
        }
    }

    return false;
}