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.Scanner

fun main(args: Array<String>) {
  val scanner = Scanner(System.`in`)
  val n = Integer.parseInt(scanner.next())
  val m = Integer.parseInt(scanner.next())
  val k = Integer.parseInt(scanner.next())
  val uf = UnionFind(n)
  val ex = Array<MutableList<Int>>(n) { mutableListOf() }
  repeat(m) {
    val a = Integer.parseInt(scanner.next())-1
    val b = Integer.parseInt(scanner.next())-1
    uf.unite(a, b)
    ex[a].add(b)
    ex[b].add(a)
  }
  repeat(k) {
    val c = Integer.parseInt(scanner.next())-1
    val d = Integer.parseInt(scanner.next())-1
    ex[c].add(d)
    ex[d].add(c)
  }
  var ans = ""
  for (i in 0 until n) {
    var x = uf.size(i)-1
    for (j in ex[i]) {
      if (uf.same(i, j)) x--
    }
    ans += x.toString() + if (i == n-1) "\n" else " "
  }
  println(ans)
}

class UnionFind(private val n: Int) {
  private val parent = IntArray(n) { it }
  private val rank = IntArray(n)
  private val size = IntArray(n) { 1 }
  
  fun find(i: Int): Int {
    if (parent[i] != i) parent[i] = find(parent[i])
    return parent[i]
  }
  
  fun same(i: Int, j: Int) = find(i) == find(j)
  
  fun unite(i: Int, j: Int) {
    val x = find(i)
    val y = find(j)
    if (x == y) return
    if (rank[x] < rank[y]) {
      parent[x] = y
      size[y] += size[x]
    } else {
      parent[y] = x
      size[x] += size[y]
      if (rank[x] == rank[y]) rank[x]++
    }
  }
  
  fun size(i: Int) = size[find(i)]
}