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: Accepted)

import java.util.*

val MAX_N = 100000
val par = Array(MAX_N) {0}
val rank = Array(MAX_N) {0}
val size = Array(MAX_N) {0}

fun init() {
    for(i in 0 until MAX_N) {
        par[i] = i
        rank[i] = 0
        size[i] = 1
    }
}

fun find(a: Int): Int {
    if (par[a] == a){
        return a
    } else{
        par[a] = find(par[a])
        return par[a]
    }
}

fun unite(a:Int, b: Int) {
    val ta = find(a)
    val tb = find(b)
    if(ta == tb) return

    if(rank[ta] < rank[tb]){
        par[ta] = tb
        size[tb] += size[ta]
    }else{
        par[tb] = ta
        size[ta] += size[tb]
        if(rank[ta] == rank[tb]) rank[ta]++;
    }
}

fun same(a: Int, b: Int): Boolean{
    return find(a) == find(b);
}

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val m = sc.nextInt()
    val k = sc.nextInt()
    init()
    val countAB = Array(n) {0}
    repeat(m) {
        val a = sc.nextInt() - 1
        val b = sc.nextInt() - 1
        unite(a, b)
        countAB[a]++
        countAB[b]++
    }
    val countCD = Array(n) { mutableSetOf<Int>()}
    repeat(k) {
        val c = sc.nextInt() - 1
        val d = sc.nextInt() - 1
        countCD[c].add(d)
        countCD[d].add(c)
    }

    println((0 until n).map {
        val blockCount = countCD[it].count { pos -> same(it, pos) }
        (size[find(it)] - 1) - countAB[it] - blockCount
    }.joinToString(" "))
}