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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import io.StdIn._
import scala.collection.mutable
 
object Main extends App {
  val firstLine = readLine.split(" ").map(_.toInt)
  val N = firstLine(0)
  val M = firstLine(1)
  val K = firstLine(2)
 
  val friends = Array.fill(N + 1)(Array.empty[Int])
  val blocks = Array.fill(N + 1)(Array.empty[Int])
 
  for (_ <- 0 until M) {
    val friendship = readLine.split(" ").map(_.toInt)
    val a = friendship(0);
    val b = friendship(1);
    friends(a) :+= b
    friends(b) :+= a
  }
 
  for (_ <- 0 until K) {
    val blockship = readLine.split(" ").map(_.toInt)
    val a = blockship(0)
    val b = blockship(1)
    blocks(a) :+= b
    blocks(b) :+= a
  }
 
  val groupNumbered = Array.fill(N + 1)(-1)
  val checked = mutable.Set.empty[Int]
 
  for (i <- 1 to N) {
    if (!checked.contains(i)) {
      groupNumbered(i) = i
      val stack = mutable.Stack[Int](friends(i):_*)
      while (stack.nonEmpty) {
        val next = stack.pop
        groupNumbered(next) = i
        checked.add(next)
        stack.pushAll(friends(next).filterNot(checked.contains))
      }
      checked.add(i)
    }
  }
 
  for (i <- 1 to N) {
    val groupId = groupNumbered(i)
 
    val count = groupNumbered.zipWithIndex.count { case (gid, id) =>
      gid == groupId && // 同じグループ
      id != i && // 自分自身ではなく
      !blocks(i).contains(id) && // ブロック関係でなく
      !friends(i).contains(id) // 直接の友人関係でもない
    }
 
    print(s"$count ")
  }
 
  print("\n")
}