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: Accepted)

def get_i() #空白区切の入力を数値(整数)の配列で返す
  return gets.chomp.split(" ").map(&:to_i)
end
def get_f() #空白区切の入力を数値(実数)の配列で返す
  return gets.chomp.split(" ").map(&:to_f)
end
def get() #空白区切の入力を文字列の配列で返す
  return gets.chomp.split(" ")
end
def get_nsp() #入力されたものを一文字ずつに区切った文字列の配列で返す
  return gets.chomp.split("")
end
def yn_judge(bool,y="Yes",n="No") #boolに真偽を投げることで、trueならy、falseならnの値を出力する 
  return bool ? y : n 
end 
def array(size,n=1,init=nil) #nに配列の次元数、sizeに配列の大きさ、initに初期値を投げることでその配列を返す 
  if n==1
    return Array.new(size){init}
  else
    return Array.new(n).map{Array.new(size){init}}
  end
end

class UnionFindTree
  attr_accessor:par,:rank,:size
  def initialize(n)
    @par=Array.new(n){|i| i}
    @rank=array(n,1,0)
    @size=array(n,1,1)
  end

  def find(x)
    return x if @par[x]==x
    @par[x]=find(@par[x])
    return @par[x]
  end

  def union(x,y)
    x=find(x)
    y=find(y)
    return if x==y
    if @rank[x]<@rank[y]
      @par[x]=y
      @size[y]+=@size[x]
    else
      @par[y]=x
      @size[x]+=@size[y]
      @rank[x]+=1 if @rank[x]==@rank[y]
    end
  end

  def same?(x,y)
    return find(x)==find(y)
  end

  def size(x)
    return @size[find(@par[x])]
  end
end

n,m,k=get_i
uft=UnionFindTree.new(n)
friends=array(n,1,0)
blocks=array(n,1,0)
ans=[]
m.times do|i|
  a,b=get_i
  a-=1
  b-=1
  uft.union(a,b)
  friends[a]+=1
  friends[b]+=1
end
k.times do|i|
  c,d=get_i
  c-=1
  d-=1
  #uft.union(c,d)
  if uft.same?(c,d)
    blocks[c]+=1
    blocks[d]+=1
  end
end
n.times do|i|
  ans.push(uft.size(i)-1-friends[i]-blocks[i])
end
puts ans.join(" ")