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)

fun main(args: Array<String>) {
    val (N, M, K) = listOfInt()
    val U = UnionFind(N)
    val F = Array<MutableList<Int>>(N) { mutableListOf<Int>() }
    val G = Array<MutableList<Int>>(N) { mutableListOf<Int>() }
    repeat(M) {
        val (a, b) = listOfInt(-1)
        U.union(a, b); U.union(b, a)
        F[a].add(b); F[b].add(a)
    }
    repeat(K) {
        val (c, d) = listOfInt(-1)
        G[c].add(d)
        G[d].add(c)
    }
    val ans = IntArray(N)
    repeat(N) { n ->
        var f = 0
        repeat(N) { m ->
            if (n != m) {
                if (!F[n].contains(m) && !G[n].contains(m)) {
                    if (U.find(n) == U.find(m)) f += 1
                }
            }
        }
        ans[n] = f
    }

    println(ans.joinToString(separator = " "))
}
fun next() = readLine()!!
fun nextInt() = Integer.parseInt(next())
fun listOfString() = next().split(" ")
fun listOfInt(delta: Int = 0) = listOfString().map { Integer.parseInt(it) + delta }
class UnionFind(n: Int) {
    var rank: IntArray
    var par:  IntArray
    var size: IntArray

    init {
        this.rank = IntArray(n) { 0 }
        this.par  = IntArray(n) { it }
        this.size = IntArray(n) { 1 }
    }

    fun union(x: Int, y: Int): Boolean {
        var rx = this.root(x)
        var ry = this.root(y)
        if (rx == ry) return false

        if (this.rank[rx] < this.rank[ry]) {
            val w = rx.toInt()
            rx = ry
            ry = w
        } else if (this.rank[rx] == this.rank[ry]) {
            this.rank[rx] += 1
        }
        this.par[ry] = rx.toInt()
        this.size[rx] += this.size[ry]
        return true
    }

    fun find(x: Int): Int {
        if (this.par[x] == x) return x
        else {
            this.par[x] = this.find(this.par[x])
            return this.par[x]
        }
    }
    fun isSame(x: Int, y: Int): Boolean = this.root(x) == this.root(y)
    fun getSize(x: Int): Int = this.size[this.find(x)]
    fun root(x: Int): Int = this.find(x)
    fun unite(x: Int, y: Int): Boolean = this.union(x, y)

}