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: Runtime Error)

import java.util.*

fun main(args : Array<String>) {
    val (n, m, k) = readLine()!!.split(" ").map { it.toInt() }
    val friendLists = Array(n) { mutableListOf<Int>() }
    val blockLists = Array(n) { BooleanArray(n) {false} }
    val networks = Array(n) { mutableListOf<Int>() }

    repeat(m) {
        val (a, b) = readLine()!!.split(" ").map { it.toInt() }
        friendLists[a-1].add(b-1)
        friendLists[b-1].add(a-1)
    }

    repeat(k) {
        val (c, d) = readLine()!!.split(" ").map { it.toInt() }
        blockLists[c-1][d-1] = true
        blockLists[d-1][c-1] = true
    }

    val suggestions = IntArray(n) { 0 }

    repeat(n) {
        if (networks[it].size > 0) {
            val blocked = networks[it].filter { idx -> blockLists[it][idx] }.size
            suggestions[it] = networks[it].size - friendLists[it].size - blocked - 1
            return@repeat
        }

        var cnt = 0
        val stack = Stack<Int>()
        stack.push(it)
        val seen = BooleanArray(n) {false}
        seen[it] = true
        val members = mutableListOf<Int>()

        while(stack.isNotEmpty()) {
            cnt++
            val idx = stack.pop()
            val friends = friendLists[idx]
            members.add(idx)

            friends.forEach { idx ->
                if (!seen[idx]) {
                    stack.push(idx)
                    seen[idx] = true
                }
            }
        }

        members.forEach { idx ->
            networks[idx] = members
        }

        val blocked = networks[it].filter { idx -> blockLists[it][idx] }.size
        suggestions[it] = Math.max(networks[it].size - friendLists[it].size - blocked - 1, 0)
    }

    println(suggestions.joinToString(separator = " "))
}