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"
)

func main() {
	sc.Split(bufio.ScanWords)
	n, m, k := nextInt(), nextInt(), nextInt()

	uf := newUnionFind(n)

	friends := make([]int, n)
	for i := 0; i < m; i++ {
		a, b := nextInt()-1, nextInt()-1
		uf.unite(a, b)
		friends[a]++
		friends[b]++
	}

	blocks := make([]int, n)
	for i := 0; i < k; i++ {
		c, d := nextInt()-1, nextInt()-1
		if uf.same(c, d) {
			blocks[c]++
			blocks[d]++
		}
	}

	for i := 0; i < n; i++ {
		cnt := uf.size(i) - friends[i] - blocks[i] - 1
		fmt.Printf("%d ", cnt)
	}
	fmt.Println()
}

var sc = bufio.NewScanner(os.Stdin)

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

// UnionFind .
type UnionFind struct {
	par   []int
	rank  []int
	sizes []int
}

func newUnionFind(n int) UnionFind {
	u := UnionFind{}
	u.par = make([]int, n)
	u.rank = make([]int, n)
	u.sizes = make([]int, n)
	for i := 0; i < n; i++ {
		u.par[i] = i
		u.rank[i] = 0
		u.sizes[i] = 1
	}
	return u
}

func (u UnionFind) unite(x, y int) {
	x = u.root(x)
	y = u.root(y)
	if x == y {
		return
	}

	if u.rank[x] < u.rank[y] {
		u.par[x] = y
		u.sizes[y] += u.sizes[x]
	} else {
		u.par[y] = x
		u.sizes[x] += u.sizes[y]
		if u.rank[x] == u.rank[y] {
			u.rank[x]++
		}
	}
}

func (u UnionFind) root(x int) int {
	if u.par[x] == x {
		return x
	}
	u.par[x] = u.root(u.par[x])
	return u.par[x]
}

func (u UnionFind) same(x, y int) bool {
	return u.root(x) == u.root(y)
}

func (u UnionFind) size(x int) int {
	return u.sizes[u.root(x)]
}