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)

fun main(args : Array<String>) {
    val (n, m, k) = readLine()!!.split(" ").map { it.toInt() }
    val excluded = Array(n) { mutableListOf<Int>() }
    val uf = UnionFind(n)

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

    repeat(k) {
        val (c, d) = readLine()!!.split(" ").map { it.toInt()-1 }
        if (uf.isSame(c, d)) {
            excluded[c].add(d)
            excluded[d].add(c)
        }
    }

    println((0 until n).map { uf.size(it) - excluded[it].size - 1 }.joinToString(separator = " "))
}

private data class UnionFind(val initSize: Int) {
    private val parents = IntArray(this.initSize) { it }
    private val sizes = LongArray(this.initSize) { 1L }

    fun root(idx: Int): Int {
        if (idx == parents[idx]) return idx

        val rootP = root(parents[idx])
        parents[idx] = rootP
        return rootP
    }

    // NOTE: merge to the larger group
    fun merge(p1: Int, p2: Int) {
        if (this.isSame(p1, p2)) return

        val rootP1 = root(p1)
        val rootP2 = root(p2)

        if (this.sizes[rootP1] >= this.sizes[rootP2]) {
            this.sizes[rootP1] += this.sizes[rootP2]
            this.parents[rootP2] = this.parents[rootP1]
        } else {
            this.sizes[rootP2] += this.sizes[rootP1]
            this.parents[rootP1] = this.parents[rootP2]
        }
    }

    fun isSame(p1: Int, p2: Int): Boolean {
        return this.root(p1) == this.root(p2)
    }

    fun size(idx: Int): Long {
        return this.sizes[this.root(idx)]
    }
}