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)

private fun readAtOnce(): List<String> = readLine()!!.split(" ")
private fun readInts(): List<Int> = readAtOnce().map(String::toInt)
private fun readLongs(): List<Long> = readAtOnce().map(String::toLong)
private fun zeroArray(size: Int): IntArray = IntArray(size) { 0 }
private fun Int.repeat(predicate: (Int) -> Unit) = repeat(this, predicate)

private class UnionFind(N: Int) {
    private val parent = Array(N) { -1 }
    private val rank = zeroArray(N)

    fun findRoot(x: Int): Int =
            if (parent[x] < 0) {
                x
            } else {
                parent[x] = findRoot(parent[x])
                parent[x]
            }

    fun unite(x: Int, y: Int) {
        val rootX = findRoot(x)
        val rootY = findRoot(y)
        if (rootX == rootY) return

        if (rank[rootX] < rank[rootY]) {
            parent[rootY] += parent[rootX]
            parent[rootX] = rootY
        } else {
            parent[rootX] += parent[rootY]
            parent[rootY] = rootX
            if (rank[rootX] == rank[rootY]) rank[rootX]++
        }
    }

    fun hasInSameGroup(x: Int, y: Int): Boolean = findRoot(x) == findRoot(y)

    fun groupSizeOf(belongs: Int) = -parent[findRoot(belongs)]
}

// heap
private val friends = zeroArray(100004)
private val blocking = Array(100004) { mutableListOf<Int>() }

fun main(args: Array<out String>) {
    val (N, M, K) = readInts()
    val unionFind = UnionFind(N + 1)
    M.repeat {
        val (A, B) = readInts()
        unionFind.unite(A, B)
        friends[A]++
        friends[B]++
    }
    K.repeat {
        val (C, D) = readInts()
        blocking[C].add(D)
        blocking[D].add(C)
    }

    (1..N)
            .map { n -> unionFind.groupSizeOf(n) - 1 - friends[n] - blocking[n].count { blocked -> unionFind.hasInSameGroup(blocked, n) } }
            .joinToString(separator = " ")
            .let(::println)
}