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: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
  val (N, M, K) = readInts()
  val AB = (1..M).map { readInts() }
  val CD = (1..K).map { readInts() }

  // ブロックメモ作成
  val blocks = Array(N) { HashSet<Int>() }
  CD.forEach {
    val (C, D) = it
    blocks[C - 1].add(D - 1)
    blocks[D - 1].add(C - 1)
  }

  // 友達グラフ作成
  val G = Array(N) { HashSet<Int>() }
  AB.forEach {
    val (A, B) = it
    G[A - 1].add(B - 1)
    G[B - 1].add(A - 1)
  }

  // グループ作成
  val groups = Array(N) { emptySet<Int>() }
  for (i in 0 until N) {
    if (groups[i].isNotEmpty()) continue
    val set = HashSet<Int>()
    fun f(n: Int) {
      if (groups[n].isNotEmpty()) return
      set.add(n)
      groups[n] = set
      G[n].forEach { f(it) }
    }
    f(i)
  }

  // 友達候補の出力
  for (i in 0 until N) {
    // 所属グループ - 既存友達 - ブロック者 - 自分 が候補になる
    val group = groups[i]
    val friends = G[i]
    val block = blocks[i]

    val friendCnt = friends.count { group.contains(it) } // 所属グループ内の既存友達の数
    val blockCnt = block.count { group.contains(it) } // 所属グループ内のブロック者の数
    if (i != 0) print(" ")
    print(group.size - friendCnt - blockCnt - 1)
  }
  println()
}

fun readString() = readLine()!!
fun readStrings() = readString().split(" ")
fun readInt() = readString().toInt()
fun readInts() = readStrings().map { it.toInt() }
fun readLong() = readString().toLong()
fun readLongs() = readStrings().map { it.toLong() }