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"

var N, M, K int

type Unionfind struct {
	par   []int
	siz []int
}

func (u *Unionfind) Init(n int) {
	u.par = make([]int, n)
	u.siz = make([]int, n)
	for i := range u.par {
		u.par[i] = i
		u.siz[i] = 1
	}
}

func (u *Unionfind) Root(x int) int {
	if u.par[x] == x {
		return x
	}
	rx := u.Root(u.par[x])
	u.par[x] = rx
	return u.par[x]
}
func (u *Unionfind) Same(x, y int) bool {
	rx, ry := u.Root(x), u.Root(y)
	return rx == ry
}
func (u *Unionfind) Unite(x, y int) {
	rx, ry := u.Root(x), u.Root(y)
	if rx == ry {
		return
	}
	if u.siz[ry] < u.siz[rx] {
		rx, ry = ry, rx
	}
	u.siz[ry] += u.siz[rx]
	u.par[rx] = ry
}
func (u *Unionfind) Size(x int) int {
	return u.siz[u.Root(x)]
}

func main() {
	fmt.Scan(&N, &M, &K)
	var fs, bs [][]int
	var funion Unionfind
	funion.Init(N)
	fs = make([][]int, N)
	bs = make([][]int, N)

	for i := 0; i < M; i++ {
		var a, b int
		fmt.Scan(&a, &b)
		a, b = a-1, b-1
		fs[a] = append(fs[a], b)
		fs[b] = append(fs[b], a)
		funion.Unite(a, b)
	}
	for i := 0; i < K; i++ {
		var a, b int
		fmt.Scan(&a, &b)
		a, b = a-1, b-1
		bs[a] = append(bs[a], b)
		bs[b] = append(bs[b], a)
	}

	cs := make([]int, N)
	for a := 0; a < N; a++ {
		siz := funion.Size(a) -1
		for _, b := range fs[a] {
			if funion.Same(a, b) {
				siz--
			}
		}
		for _, b := range bs[a] {
			if funion.Same(a, b) {
				siz--
			}
		}
		cs[a] = siz
	}

	for i, c := range cs {
		if i == len(cs)-1 {
			fmt.Printf("%d\n", c)
		} else {
			fmt.Printf("%d ", c)
		}
	}
}

func contains(as []int, b int) bool {
	for _, a := range as {
		if a == b {
			return true
		}
	}
	return false
}