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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class UnionFindTree
	#0→連結成分のサイズ 1→parent 2→rank 3→連結成分に含まれる人
	attr_accessor :nodes
	def initialize(size)
		@nodes = Array.new(size)
		size.times do |i|
			@nodes[i] = [1, i, 0, [i]]
		end
		nodes = @nodes
	end

	# x番目の値の親を探す
	def findParent(x)
		if @nodes[x][1] == x then 
			return x
		else
			return findParent(@nodes[x][1])
		end
	end

	# 合併
	def unite(x, y)
		xx = findParent(x)
		yy = findParent(y)
		return if xx == yy

		xx, yy = yy, xx if @nodes[xx][2] > @nodes[yy][2]

		@nodes[xx][0] += @nodes[yy][0]
		@nodes[yy][1] = xx
		@nodes[xx][2] += 1 if @nodes[xx][2] == @nodes[y][2]
		#@nodes[xx][3] += @nodes[yy][3]

		# if @nodes[xx][2] < @nodes[yy][2] then
		# 	@nodes[xx][1] = yy
		# 	@nodes[xx][0] += @nodes[yy][0]
		# 	@nodes[yy][0] = @nodes[xx][0]
		# else
		# 	@nodes[yy][1] = xx
		# 	@nodes[yy][0] += @nodes[xx][0]
		# 	@nodes[xx][0] = @nodes[yy][0]
		# 	@nodes[xx][2] += 1 if @nodes[xx][2] == @nodes[yy][2]
		# end
	end

	def same(x, y)
		return findParent(x) == findParent(y)
	end
end

n, m, k = gets.chomp.split(" ").map(&:to_i)
group = UnionFindTree.new(n)
kankei = Array.new(n).map{Array.new}

# p group.nodes
m.times do 
	a, b = gets.chomp.split(" ").map{|ngo| ngo.to_i - 1}
	group.unite(a,b)
	#p group.nodes.map{|ma| ma[1]}
	# p group.nodes
 	kankei[a] << b
	kankei[b] << a
end

k.times do
	a, b = gets.chomp.split(" ").map{|ngo| ngo.to_i - 1}
	kankei[a] << b
	kankei[b] << a
end

yaju = []
# puts "group"; p group.nodes ;puts 
# puts "kankei"; p kankei; puts
n.times do |i|
	par = group.findParent(i)
	num = group.nodes[par][0]
	unchi = kankei[i].count{|hotti| group.findParent(hotti) == par}
	yaju << num - unchi - 1
	#puts "par:#{par} num:#{num} unchi:#{unchi} yaju:#{yaju}"
	#puts "i:#{i} par:#{par} num:#{group.nodes[par][0]} group:#{group.nodes[i][3]}"
	# list = group.nodes[par][3]
	# ngo = list - hash[i]
	#print ((group.nodes[par][3] - hash[i]).length - 1).to_s + " "
end
puts yaju.join(" ")