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.concurrent.LinkedBlockingDeque

fun main(args: Array<String>) {
    val (n, m, k) = readLine()?.split(" ")?.map(String::toInt) ?: return
    class Person(
        val name: Int
    ) {
        val friends = hashSetOf<Int>()
        val blocks = hashSetOf<Int>()
    }

    val people = hashMapOf<Int, Person>()
    repeat(m) {
        val (a, b) = readLine()?.split(" ")?.map(String::toInt) ?: return
        people[a]?.friends?.add(b) ?: run {
            people[a] = Person(a).apply {
                friends.add(b)
            }
        }
        people[b]?.friends?.add(a) ?: run {
            people[b] = Person(b).apply {
                friends.add(a)
            }
        }
    }
    repeat(k) {
        val (a, b) = readLine()?.split(" ")?.map(String::toInt) ?: return
        people[a]?.blocks?.add(b) ?: run {
            people[a] = Person(a).apply {
                blocks.add(b)
            }
        }
        people[b]?.blocks?.add(a) ?: run {
            people[b] = Person(b).apply {
                blocks.add(a)
            }
        }
    }
    val suggestions = hashMapOf<Int, Int>()
    0.until(n).forEach {
        suggestions[it + 1] = 0
    }
    val remain = people.keys.toMutableSet()
    while (remain.isNotEmpty()) {
        val queue = LinkedBlockingDeque<Int>()
        val cluster = hashMapOf<Int, Person>()
        queue.offer(remain.first())
        while (queue.isNotEmpty()) {
            val p = queue.poll()
            cluster[p] = people[p]!!
            people[p]!!.friends.forEach { friend ->
                if (!cluster.containsKey(friend)) {
                    cluster[friend] = people[friend]!!.apply {
                        this.friends.filter {
                            !cluster.containsKey(it)
                        }.forEach {
                            queue.offer(it)
                        }
                    }
                }
            }
        }
        cluster.values.map {
            val diff = (it.blocks - cluster.keys).size
            suggestions[it.name] = cluster.keys.size - (1 + it.friends.size + it.blocks.size - diff)
        }
        remain.removeAll(cluster.keys)
    }
    val output = suggestions.keys.sorted().map {
        suggestions[it]
    }.joinToString(" ")

    println(output)
}