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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

N,M,K = list(map(int,input().split()))
import sys
input = sys.stdin.readline
from collections import defaultdict
class UF():
  def __init__(self,n):
    self.n_node = n
    self.l = [i for i in range(n)]
    self.size = [1 for i in range(n)]
    self.d = None
  
  def union(self,i,j):
    if not self.connected(i,j):
      iroot,ipath = self.root(i,return_path=True)
      jroot,jpath = self.root(j,return_path=True)
      if len(jpath) < len(ipath):
        for node in ipath + jpath:
          self.l[node] = iroot
      else:
        for node in ipath + jpath:
          self.l[node] = jroot
        
  def root(self,i,return_path=False):
    path = [i]
    while self.l[i] != i:
      i = self.l[i]
      path.append(self.l[i])
    
    if return_path:
      return i,path
    else:
      return i
  
  def connected(self,i,j):
    if self.root(i) == self.root(j):
      return True
    else:
      return False
  
  def simplify(self):
    for node in range(self.n_node):
      root = self.root(self.l[node])
      if self.l[node] != root:
        self.l[node] = root
  
  def calc_connected(self):
    self.d = defaultdict(set)
    self.simplify()
    for idx,root in enumerate(self.l):
      self.d[root] |= {idx}
    
  def connected_set(self,i):
    if self.d is None:
      self.calc_connected()
    return self.d[self.root(i)]
      
uf = UF(N)
direct_connected = defaultdict(set)
for i in range(M):
  i,j = map(int,input().split())
  i-=1
  j-=1
  
  uf.union(i,j)
  direct_connected[i] |= {j}
  direct_connected[j] |= {i}

block_list = defaultdict(set)
for i in range(K):
  i,j = map(int,input().split())
  i-=1
  j-=1
  block_list[i] |={j}
  block_list[j] |={i}

  
uf.calc_connected()
for i in range(N):
  print(len(uf.d[uf.l[i]] - direct_connected[i] - block_list[i]-{i}),end=" ")