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: Accepted)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n = sc.nextInt()
  val m = sc.nextInt()
  val k = sc.nextInt()
  val uf = UnionFind(n)
  val ans = new Array[Int](n)
  for (i <- 0 until m) {
    val a = sc.nextInt() - 1
    val b = sc.nextInt() - 1
    uf.unite(a, b)
    ans(a) = ans(a) - 1
    ans(b) = ans(b) - 1
  }
  for (i <- 0 until n) {
    ans(i) = ans(i) + uf.size(i) - 1
  }
  for (i <- 0 until k) {
    val c = sc.nextInt() - 1
    val d = sc.nextInt() - 1
    if (uf.same(c, d)) {
      ans(c) = ans(c) - 1
      ans(d) = ans(d) - 1
    }
  }
  for (i <- 0 until n) {
    print(ans(i))
    if (i < n - 1) print(" ")
  }
  println()
  case class UnionFind(val n: Int) {
    val par = Array.fill(n)(-1)
    val sz = Array.fill(n)(1)
    def find(x: Int): Int = {
      if (par(x) == -1) par(x) = x
      if (par(x) != x) {
        par(x) = find(par(x))
      }
      par(x)
    }
    def unite(x: Int, y: Int): Unit = {
      val a = find(x)
      val b = find(y)
      if (a == b) return
      if (sz(a) > sz(b)) {
        sz(a) += sz(b)
        par(b) = par(a)
      } else {
        sz(b) += sz(a)
        par(a) = par(b)
      }
    }
    def same(x: Int, y: Int): Boolean = {
      find(x) == find(y)
    }
    def size(x: Int): Int = {
      sz(find(x))
    }
  }
}