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 (1.14.1) (judge verdict: Wrong Answer)

package main

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

var sc *bufio.Scanner
var out *bufio.Writer

const INF = 1 << 30

func main() {
	sc = bufio.NewScanner(os.Stdin)
	//bytes, _ := ioutil.ReadFile("./stdin.txt")
	//sc = bufio.NewScanner(strings.NewReader(string(bytes)))
	out = bufio.NewWriter(os.Stdout)
	buf := make([]byte, 1<<20)
	sc.Buffer(buf, bufio.MaxScanTokenSize)
	sc.Split(bufio.ScanWords)
	defer out.Flush()

	N, M, K := nextInt3()
	G1 := make([][]int, N+1)
	G2 := make([][]int, N+1)
	for i := 0; i < M; i++ {
		a, b := nextInt2()
		G1[a] = append(G1[a], b)
		G1[b] = append(G1[b], a)
	}
	for i := 0; i < K; i++ {
		a, b := nextInt2()
		G2[a] = append(G2[a], b)
		G2[b] = append(G2[b], a)
	}
	G := [2]*[][]int{&G1, &G2}
	group := [2][]int{make([]int, N+1), make([]int, N+1)}
	for i := 0; i < 2; i++ {
		id := 1
		for j := 1; j <= N; j++ {
			if group[i][j] == 0 {
				dfs(*G[i], group[i], j, id)
				id++
			}
		}
	}
	c := make(map[int][]int)
	for i := 1; i <= N; i++ {
		c[group[0][i]] = append(c[group[0][i]], i)
	}
	for i := 1; i <= N; i++ {
		result := -len(G1[i])
		for _, user := range c[group[0][i]] {
			if user == i {
				continue
			}
			if group[1][user] == group[1][i] {
				continue
			}
			result++
		}
		fmt.Fprintf(out, "%d", result)
		if i == N {
			fmt.Fprintln(out, "")
		} else {
			fmt.Fprint(out, " ")
		}
	}

}

func dfs(G [][]int, group []int, u, id int) {
	group[u] = id
	for e := 0; e < len(G[u]); e++ {
		v := G[u][e]
		if group[v] == 0 {
			dfs(G, group, v, id)
		}
	}
}

func next() string {
	sc.Scan()
	return sc.Text()
}
func nextInt() int {
	a, _ := strconv.Atoi(next())
	return a
}
func nextInt2() (int, int)      { return nextInt(), nextInt() }
func nextInt3() (int, int, int) { return nextInt(), nextInt(), nextInt() }
func nextInts(n int) []int {
	ret := make([]int, n)
	for i := 0; i < n; i++ {
		ret[i] = nextInt()
	}
	return ret
}
func cond(condition bool, t, f int) int {
	if condition {
		return t
	}
	return f
}
func abs(a int) int { return cond(a < 0, -a, a) }
func min(a ...int) int {
	ret := a[0]
	for i := 1; i < len(a); i++ {
		chmin(&ret, a[i])
	}
	return ret
}
func max(a ...int) int {
	ret := a[0]
	for i := 1; i < len(a); i++ {
		chmax(&ret, a[i])
	}
	return ret
}
func chmin(a *int, b int) { *a = cond(*a < b, *a, b) }
func chmax(a *int, b int) { *a = cond(*a > b, *a, b) }