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


fscanf(STDIN, "%d%d%d", $N, $M, $K);

$friend = array_fill(1, $N, []);
for ($i = 0; $i < $M; $i++) {
    $tmp = explode(" ", trim(fgets(STDIN)));
    $friend[$tmp[0]][$tmp[1]] = 1;
    $friend[$tmp[1]][$tmp[0]] = 1;
}
$block = array_fill(1, $N, []);
for ($i = 0; $i < $K; $i++) {
    $tmp = explode(" ", trim(fgets(STDIN)));
    $block[$tmp[0]][$tmp[1]] = 1;
    $block[$tmp[1]][$tmp[0]] = 1;
}

$group_no = 0;
$group_member = [];
$group_map = array_fill(1, $N, 0);
for ($i = 1; $i <= $N; $i++) {
    if ($group_map[$i] > 0) {
        continue;
    }
    $group_no++;
    $start = $i;
    $queue = [];
    $group_member[$group_no] = 1;

    $group_map[$i] = $group_no;
    if (count($friend[$start]) === 0) {
        continue;
    }
    $q = 0;
//初期キューの登録
    getQueue($friend[$start], $group_no);
    while (isset($queue[$q])) {
        getQueue($friend[$queue[$q][0]], $group_no);
        unset($queue[$q]);
        $q++;
    }
}
foreach ($group_map as $key => $value) {
    $block_num = 0;
    foreach ($block[$key] as $bkey => $bvalue) {
        if ($group_map[$bkey] === $value) {
            $block_num++;
        }
    }
    echo $group_member[$value] - 1 - $block_num - count($friend[$key]) . " ";
}


/**
 * @param array $array
 * @param int $last_vec
 * @param int $count
 * @return int
 */
function getQueue(array $array, int $group_no)
{
    global $queue, $group_map, $group_member;
    foreach ($array as $key => $value) {
        if ($group_map[$key] === 0) {
            $group_map[$key] = $group_no;

            $group_member[$group_no]++;
            $queue[] = [$key, $group_no];
        }
    }

    return 0;

}