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 Python (3.4.3) (judge verdict: Runtime Error)

class UnionFind:
  def __init__(self,n):
    self.n=n
    self.parent=[-1 for i in range(n)]
    
  def union(self,a,b):
    if self.root(a) == self.root(b):
      return
    if self.size(a) > self.size(b):
      self.parent[a-1] += self.parent[b-1]
      self.parent[b-1] = a-1
    else:
      self.parent[a-1] = b-1
      self.parent[b-1] += self.parent[a-1]
      
  def root(self,a):
    if self.parent[a-1]==-1:
      return a-1
    else:
      return self.root(self.parent[a-1])
    
  def size(self,a):
    return -self.parent[self.root(a)]
  
n,m,k=map(int,input().split())
uni=UnionFind(n)
friend=[0]*n
for i in range(m):
  a,b=map(int,input().split())
  uni.union(a,b)
  friend[a-1]+=1
  friend[b-1]+=1
block=[0]*n
for i in range(k):
  a,b=map(int,input().split())
  if uni.root(a)==uni.root(b):
    block[a]+=1
    block[b]+=1
ans=[0]*n
for i in range(n):
  ans[i]=uni.size(i)-friend[i]-block[i]-1
for i in range(n):
  print(max(0,ans[i]),end=' ')
else:
  print()