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: Time Limit Exceeded)

import sys
input = sys.stdin.readline
 
N,M,K = list(map(int,input().split()))
ab = [list(map(int,input().split())) for _ in range(M)]
cd = [list(map(int,input().split())) for _ in range(K)]

class UnionADD():
  def __init__(self, N):
    self.N = N
    self.f = [1<<self.N]*self.N
    self.b = [1<<self.N]*self.N
    self.fb = [1<<self.N]*self.N
  
  def union(self, xx):
    t = [1<<self.N]*self.N
    for a,b in xx:
      a -= 1
      b -= 1
      t[a] = t[a] | 1<<(self.N-b-1)
      t[b] = t[b] | 1<<(self.N-a-1)
    return t
  
  def conv(self, ab,cd):
    self.f = self.union(ab)
    self.b = self.union(cd)
    self.fb = [x | y for x,y in zip(self.f, self.b)]
  
  def add_link(self, ans):
      for p,x in enumerate(ans):
        for i,y in enumerate(str(bin(x)[3:])):
          if y == '1':
            ans[p] = ans[p] | self.f[i]
      return ans
  
  def link(self,ans):
    pre = []
    ans = self.f.copy()
    while ans != pre:
      pre = ans.copy()
      ans = self.add_link(ans)
    return ans
  
  def count_link(self, ANS):
    ans = []
    for p in range(self.N):
      ANS[p] = ANS[p] | self.fb[p]
      ANS[p] = ANS[p] ^ self.fb[p]
      ans += [bin(ANS[p])[3:].count('1')]
    return ans
    

ua = UnionADD(N)
ua.conv(ab,cd)
ans = [1<<N]*N

ANS = ua.link(ans)
ans = ua.count_link(ANS)

print(*ans)