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"
	"os"
	"strconv"
	"strings"
)

// input scanner
var sc = bufio.NewScanner(os.Stdin)

const (
	initialBufSize = 10e4
	maxBufSize     = 10e6
)

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

func nextInt() int {
	a, _ := strconv.Atoi(next())
	return a
}

func nextInt64() int64 {
	a, _ := strconv.ParseInt(next(), 10, 64)
	return a
}

func nextFloat64() float64 {
	a, _ := strconv.ParseFloat(next(), 64)
	return a
}

func nextInts(n int) []int {
	ret := make([]int, n)
	for i := 0; i < n; i++ {
		ret[i] = nextInt()
	}
	return ret
}
func nextFloats(n int) []float64 {
	ret := make([]float64, n)
	for i := 0; i < n; i++ {
		ret[i] = nextFloat64()
	}
	return ret
}
func nextStrings(n int) []string {
	ret := make([]string, n)
	for i := 0; i < n; i++ {
		ret[i] = next()
	}
	return ret
}

// util
func max(l []int) (int, int) {
	idx := 0
	ans := l[idx]
	for i, n := range l {
		if n > ans {
			idx = i
			ans = n
		}
	}
	return idx, ans
}

func min(l []int) (int, int) {
	idx := 0
	ans := l[idx]
	for i, n := range l {
		if n < ans {
			idx = i
			ans = n
		}
	}
	return idx, ans
}

func sameUnion(g [][]int, a int) map[int]struct{} {
	set := make(map[int]struct{}, len(g))
	st := make([]int, 0, len(g))
	st = append(st, a)
	for len(st) > 0 {
		current := st[0]
		st = st[1:]
		set[current] = struct{}{}
		for _, next := range g[current] {
			_, exist := set[next]
			if !exist {
				st = append([]int{next}, st...)
			}
		}
	}
	return set
}

// main
func main() {
	buf := make([]byte, initialBufSize)
	sc.Buffer(buf, maxBufSize)
	sc.Split(bufio.ScanWords)
	n := nextInt()
	g := make([][]int, n)
	bg := make([][]int, n)
	for i := 0; i < n; i++ {
		g[i] = make([]int, 0, n)
		bg[i] = make([]int, 0, n)
	}
	m := nextInt()
	k := nextInt()
	for i := 0; i < m; i++ {
		ai := nextInt()
		bi := nextInt()
		g[ai-1] = append(g[ai-1], bi-1)
		g[bi-1] = append(g[bi-1], ai-1)
	}
	for i := 0; i < k; i++ {
		ci := nextInt()
		di := nextInt()
		bg[ci-1] = append(bg[ci-1], di-1)
		bg[di-1] = append(bg[di-1], ci-1)
	}
	// fmt.Println(g)
	// fmt.Println(bg)
	ans := make([]string, n)
	for i := 0; i < n; i++ {
		u := sameUnion(g, i)
		// fmt.Println(u)
		var fcount int
		for _, x := range g[i] {
			_, exist := u[x]
			if exist {
				fcount++
			}
		}
		var bcount int
		for _, x := range bg[i] {
			_, exist := u[x]
			if exist {
				bcount++
			}
		}
		ans[i] = strconv.Itoa(len(u) - fcount - bcount - 1)
	}
	fmt.Println(strings.Join(ans, " "))
}