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: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var sc = bufio.NewScanner(os.Stdin)

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

var searched []bool
var count int

func main() {
	sc.Split(bufio.ScanWords)
	n, _ := strconv.Atoi(stringLineScan())
	m, _ := strconv.Atoi(stringLineScan())
	k, _ := strconv.Atoi(stringLineScan())
	f := make(map[int][]int)
	block := make(map[int][]int)
	ab := make(map[string]struct{})
	for i := 0; i < m; i++ {
		a, _ := strconv.Atoi(stringLineScan())
		b, _ := strconv.Atoi(stringLineScan())
		a--
		b--
		if _, ok := f[a]; ok == false {
			x := make([]int, 0)
			f[a] = x
		}
		if _, ok := f[b]; ok == false {
			x := make([]int, 0)
			f[b] = x
		}
		f[a] = append(f[a], b)
		f[b] = append(f[b], a)
		ab[fmt.Sprint(b, a)] = struct{}{}
		ab[fmt.Sprint(a, b)] = struct{}{}
	}
	for i := 0; i < n; i++ {
		searched = append(searched, false)
	}
	for i := 0; i < k; i++ {
		c, _ := strconv.Atoi(stringLineScan())
		d, _ := strconv.Atoi(stringLineScan())
		c--
		d--
		if _, ok := block[c]; ok == false {
			x := make([]int, 0)
			block[c] = x
		}
		if _, ok := block[d]; ok == false {
			x := make([]int, 0)
			block[d] = x
		}
		block[c] = append(block[c], d)
		block[d] = append(block[d], c)
		ab[fmt.Sprint(c, d)] = struct{}{}
		ab[fmt.Sprint(d, c)] = struct{}{}
	}
	group := make([]int, n)
	ans := make([]int, n)
	init := 0
	for i := 0; i < n; i++ {
		for j := init; j < n; j++ {
			if !searched[j] {
				searched[j] = true
				init = j
				count++
				break
			}
		}
		result := make([]int, 1)
		result[0] = init
		result = splitter(init, f[init], f, result)
		l := len(result) - 1
		for _, j := range result {
			ans[j] = l
			group[j] = init
		}
		if count == n {
			break
		}
	}
	for k, list := range f {
		for _, v := range list {
			if _, ok := ab[fmt.Sprint(k, v)]; group[k] == group[v] && ok {
				ans[k]--
			}
		}
	}
	for k, list := range block {
		for _, v := range list {
			if _, ok := ab[fmt.Sprint(k, v)]; group[k] == group[v] && ok {
				ans[k]--
			}
		}
	}
	fmt.Println(strings.Trim(fmt.Sprint(ans), "[]"))
}
func splitter(init int, list []int, f map[int][]int, result []int) []int {
	for _, i := range list {
		if !searched[i] {
			searched[i] = true
			count++
			result = append(result, i)
			result = splitter(i, f[i], f, result)
		}
	}
	return result
}