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: Runtime Error)

fun main(args: Array<String>){
    val (N, M, K) = readLine()!!.split(" ").map { it.toInt() }
    val uf = UnionFind(N)
    val friend = IntArray(N + 1)

    repeat(M){
        val (A, B) = readLine()!!.split(" ").map { it.toInt() }
        uf.unite(A, B)
        friend[A]++
        friend[B]++
    }

    val size = IntArray(N+1)
    for(i in size.indices){
        size[i] = uf.unionSize(i) - friend[i] - 1
    }

    repeat(K){
        val (C, D) = readLine()!!.split(" ").map { it.toInt() }
        if(uf.isSameUnion(C, D)){
            size[C]--
            size[D]--
        }
    }

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

class UnionFind(n: Int) {

    private val parent = IntArray(n + 1) { i -> i }
    private val size = IntArray(n + 1){ i -> 1 }

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

    fun unite(x: Int, y: Int) {
        val rx = root(x)
        val ry = root(y)
        if (rx != ry) {
            parent[rx] = ry
            size[ry] += size[rx]
        }
    }

    fun isSameUnion(x: Int, y: Int): Boolean {
        return root(x) == root(y)
    }

    fun unionSize(x: Int): Int{
        return size[root(x)]
    }
}