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 Array(n, m, k) = readLine.split(" ").map(_.toInt)

  val friends = Array.fill(n)(Array.empty[Int])
  for (_ <- 0 until m) {
    val Array(a, b) = readLine.split(" ").map(_.toInt - 1)
    friends(a) :+= b
    friends(b) :+= a
  }

  val blocks = Array.fill(n)(Array.empty[Int])
  for (_ <- 0 until k) {
    val Array(a, b) = readLine.split(" ").map(_.toInt - 1)
    blocks(a) :+= b
    blocks(b) :+= a
  }

  val groupToMember = mutable.Map.empty[Int, mutable.Set[Int]]
  val memberToGroup = Array.fill(n)(-1)
  val checked = mutable.Set.empty[Int]
  for (i <- 0 until n) {
    groupToMember.put(i, mutable.Set.empty)
    if (!checked.contains(i)) {
      val stack = mutable.Stack[Int](i +: friends(i):_*)
      while (stack.nonEmpty) {
        val next = stack.pop
        if (!checked.contains(next)) {
          memberToGroup(next) = i
          groupToMember(i).add(next)
          checked.add(next)
          stack.pushAll(friends(next))
        }
      }
    }
  }

  val ans = Range(0, n).map { i =>
    val gid = memberToGroup(i)
    groupToMember(gid).count(candidate =>
      !(i == candidate || friends(i).contains(candidate) || blocks(i).contains(candidate))
    )
  }

  println(ans.mkString(" "))
}