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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args : Array<String>) {
    val (n, m, k) = readLine()!!.split(' ').map(String::toInt)
    val friends = HashMap<Int, HashSet<Int>>()
    val blocks = HashMap<Int, HashSet<Int>>()
    (1..m).forEach {
        val (a, b) = readLine()!!.split(' ').map(String::toInt)
        if (friends[a] == null) friends[a] = HashSet<Int>()
        if (friends[b] == null) friends[b] = HashSet<Int>()
        friends[a]!! += b
        friends[b]!! += a
    }
    (1..k).forEach {
        val (c, d) = readLine()!!.split(' ').map(String::toInt)
        if (blocks[c] == null) blocks[c] = HashSet<Int>()
        if (blocks[d] == null) blocks[d] = HashSet<Int>()
        blocks[c]!! += d
        blocks[d]!! += c
    }
    (1..n).forEach { i ->
        val que = ArrayDeque<Int>()
        val blk = blocks[i]
        val dis = HashMap<Int, Int>()
        val dtr = BooleanArray(n + 1)
        var cnt = 0
        dis[i] = 0
        dtr[i] = true
        que += i
        while (que.isNotEmpty()) {
            val cur = que.remove()
            if (friends[cur] == null) continue
            for (fr in friends[cur]!!) {
                if (dtr[fr]) continue
                dis[fr] = dis[cur]!! + 1
                if (blk != null && blk.contains(fr)) continue
                dtr[fr] = true
                que += fr
                if (dis[fr]!! != 1) cnt++
            }
        }
        print(cnt)
        print(' ')
    }
    println()
}