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 (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

const (
	initialBufSize = 10000
	maxBufSize     = 1000000
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		sc.Split(bufio.ScanWords)
		return sc
	}()
	mod int = 1e9 + 7
)

type Graph struct {
	n     int
	edges [][]int
}

func NewGraph(n int) *Graph {
	g := &Graph{
		n:     n,
		edges: make([][]int, n),
	}
	return g
}

func (g *Graph) AddEdge(u, v int) {
	g.edges[v] = append(g.edges[v], u)
	g.edges[u] = append(g.edges[u], v)
}

func main() {
	N := getInt()
	M := getInt()
	K := getInt()

	friendGraph := NewGraph(N)

	for i := 0; i < M; i++ {
		a := getInt() - 1
		b := getInt() - 1
		friendGraph.AddEdge(a, b)
	}

	blockGraph := NewGraph(N)

	for i := 0; i < K; i++ {
		c := getInt() - 1
		d := getInt() - 1
		blockGraph.AddEdge(c, d)
	}

	var visitedEdges []map[int]struct{}
	results := make([]int, N)

	for i := 0; i < N; i++ {
		for _, visited := range visitedEdges {
			_, flag := visited[i]
			if flag {
				continue
			}
		}

		visited := map[int]struct{}{}
		dfs(i, friendGraph.edges, visited)
		visitedEdges = append(visitedEdges, visited)

		for _, visited := range visitedEdges {
			_, flag := visited[i]
			if flag {
				results[i] = len(visited) - len(friendGraph.edges[i]) - 1

				for _, node := range blockGraph.edges[i] {
					_, flag = visited[node]
					if flag {
						results[i]--
					}
				}

				continue
			}
		}
	}

	printIntArray(results)
}

func dfs(c int, edges [][]int, visited map[int]struct{}) {
	visited[c] = struct{}{}

	for _, v := range edges[c] {
		_, flag := visited[v]
		if flag {
			continue
		}
		dfs(v, edges, visited)
	}
}

func getInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func getString() string {
	sc.Scan()
	return sc.Text()
}

func abs(a int) int {
	return int(math.Abs(float64(a)))
}

func pow(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func powMod(n, p int) int {
	if p == 0 {
		return 1
	} else if p%2 == 0 {
		t := powMod(n, p/2)
		return calcMod(t * t)
	} else {
		return n * powMod(n, p-1)
	}
}

func min(nums ...int) int {
	if len(nums) == 0 {
		panic("funciton min() requires at least one argument.")
	}
	res := nums[0]
	for i := 0; i < len(nums); i++ {
		res = int(math.Min(float64(res), float64(nums[i])))
	}
	return res
}

func max(nums ...int) int {
	if len(nums) == 0 {
		panic("funciton max() requires at least one argument.")
	}
	res := nums[0]
	for i := 0; i < len(nums); i++ {
		res = int(math.Max(float64(res), float64(nums[i])))
	}
	return res
}

func strSearch(a []string, b string) bool {
	for i := 0; i < len(a); i++ {
		if a[i] == b {
			return true
		}
	}
	return false
}

func printIntArray(array []int) {
	strArray := fmt.Sprint(array)
	fmt.Println(strArray[1 : len(strArray)-1])
}

func calcMod(x int) int {
	return x % mod
}