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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"fmt"
)

// Queue is for bfs
type Queue struct {
	data []interface{}
}

// Push adds a value to queue
func (q *Queue) Push(v interface{}) {
	q.data = append(q.data, v)
}

// Pop removes the last element and return it
func (q *Queue) Pop() (interface{}, error) {
	if q.Empty() {
		return nil, fmt.Errorf("queue is empty")
	}
	v := q.data[0]
	q.data = q.data[1:]
	return v, nil
}

// Empty returns true if queue is empty
func (q *Queue) Empty() bool {
	return len(q.data) == 0
}

// NewQueue generates a new queue
func NewQueue() *Queue {
	return &Queue{make([]interface{}, 0)}
}

// Graph is a set of nodes
type Graph struct {
	Nodes []*Node
}

// Node has its value and neighbors
type Node struct {
	Val    interface{}
	Neighs []*Node
}

// Show shows a graph
func (g *Graph) Show() {
	nodes := g.Nodes
	for _, node := range nodes {
		fmt.Printf("%v: ", node.Val)
		for _, neigh := range node.Neighs {
			fmt.Printf("%v ", neigh.Val)
		}
		fmt.Println()
	}
}

func dfs(g *Graph, node *Node) map[*Node]bool {
	set := make(map[*Node]bool)
	q := NewQueue()
	q.Push(node)
	set[node] = true
	for !q.Empty() {
		src, _ := q.Pop()
		if v, ok := src.(*Node); ok {
			for _, neigh := range v.Neighs {
				if !set[neigh] {
					q.Push(neigh)
					set[neigh] = true
				}
			}
		}
	}
	return set
}

func findSets(g *Graph) map[*Node]map[*Node]bool {
	sets := make(map[*Node]map[*Node]bool)
	nodes := g.Nodes
	for _, node := range nodes {
		if _, ok := sets[node]; ok {
			continue
		}
		sets[node] = dfs(g, node)
		for e := range sets[node] {
			sets[e] = sets[node]
		}
	}
	return sets
}

type relation struct {
	from *Node
	to   *Node
}

func main() {
	var n, m, k int
	fmt.Scan(&n, &m, &k)
	people := make([]*Node, n)
	for i := 0; i < n; i++ {
		people[i] = &Node{i, make([]*Node, 0)}
	}

	for i := 0; i < m; i++ {
		var a, b int
		fmt.Scan(&a, &b)

		people[a-1].Neighs = append(people[a-1].Neighs, people[b-1])
		people[b-1].Neighs = append(people[b-1].Neighs, people[a-1])
	}

	g := &Graph{people}

	sets := findSets(g)

	blocked := make(map[*Node]int)
	for i := 0; i < k; i++ {
		var c, d int
		fmt.Scan(&c, &d)

		set := sets[people[c-1]]
		if set[people[d-1]] {
			blocked[people[c-1]]++
		}

		set2 := sets[people[d-1]]
		if set2[people[c-1]] {
			blocked[people[d-1]]++
		}
	}

	for i := 0; i < n; i++ {
		pi := people[i]
		set := sets[pi]
		fmt.Printf("%v ", len(set)-len(pi.Neighs)-blocked[pi]-1)
	}
	fmt.Println()
}