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 (1.3.71) (judge verdict: Accepted)

fun readString() = readLine()!!
fun readInt() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readIntList() = readLine()!!.split(" ").map(String::toInt)
fun readLongList() = readLine()!!.split(" ").map(String::toLong)
fun <T> Array<Array<T>>.print() {
    this.forEach {
        println(it.joinToString(" "))
    }
}

class UnionFind(n: Int) {
    private val parent: MutableList<Int> = MutableList(n) { it }
    private val rank: MutableList<Int> = MutableList(n) { 0 }
    private val _size: MutableList<Int> = MutableList(n) { 1 }

    private fun root(x: Int): Int {
        if (parent[x] == x) return x
        parent[x] = root(parent[x])

        return parent[x]
    }

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

        if (rank[rx] < rank[ry]) {
            val tmp = rx
            rx = ry
            ry = tmp
        }

        if (rank[rx] == rank[ry]) rank[rx]++

        parent[ry] = rx
        _size[rx] += _size[ry]
    }

    fun isSame(x: Int, y: Int): Boolean = root(x) == root(y)

    fun size(x: Int): Int = _size[root(x)]
}

fun main() {
    val (n, m, k) = readIntList()
    val friendUf = UnionFind(n + 1)
    val notCandidates = Array(n + 1) { mutableListOf<Int>() }

    for (i in 0 until m) {
        val (a, b) = readIntList()
        friendUf.union(a, b)
        notCandidates[a].add(b)
        notCandidates[b].add(a)
    }
    
    for (i in 0 until k) {
        val (c, d) = readIntList()
        if (friendUf.isSame(c, d)) {
            notCandidates[c].add(d)
            notCandidates[d].add(c)
        }
    }

    val friendCandidates = Array(n + 1) { 0 }
    for (i in 1..n) {
        friendCandidates[i] = friendUf.size(i) - notCandidates[i].size - 1
    }

    println(friendCandidates.drop(1).joinToString(separator = " "))
}