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: Runtime Error)

object Main extends App {
	import scala.collection.mutable.Map
	val sc = new java.util.Scanner(System.in)
	val pw = new java.io.PrintWriter(System.out)
	
	val n, m, k = sc.nextInt
	
	val f = Array.fill(n+1)(Array.fill(n+1)(true))
	
	val candidate = UnionFind(1 to n)
	
	Seq.fill(m)((sc.nextInt, sc.nextInt)) foreach { case (a, b) =>
		f(a)(b) = false
		f(b)(a) = false
		candidate.unite(a, b)
	}
	
	Seq.fill(k)((sc.nextInt, sc.nextInt)) foreach { case (c, d) =>
		f(c)(d) = false
		f(d)(c) = false
	}

	(1 to n) foreach { i =>
		pw.print( ((1 to n).filter(j => f(i)(j) && candidate.same(i, j)).size - 1).toString + " " )
	}
	pw.flush()
}

case class UnionFind[T](seq: Seq[T]) {
	import scala.collection.mutable.Map
	private[this] val par: Map[T, T] = Map[T,T](seq.map(elem => elem -> elem): _*)

	@scala.annotation.tailrec
	private[this] def find(t: T, acc: Seq[T] = Nil): T = {
		val p = par(t)
		if (p == t) {acc foreach { par(_) = t }; t} else find(p, p +: acc)
	}

	def unite(a: T, b: T): Unit = {
		val pa = find(a)
		val pb = find(b)
		if (pa != pb) par(pa) = pb
	}

	def same(a: T, b: T) = find(a) == find(b)
}