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: Wrong Answer)

import numpy as np
N,M,K = list(map(int,input().split()))
R = np.array([[0]*N]*N)

for i in range(M):
  a,b = list(map(int,input().split()))
  R[a-1][b-1] = 2*N
  R[b-1][a-1] = 2*N

for i in range(K):
  c,d = list(map(int,input().split()))
  R[c-1][d-1] = -2*N
  R[d-1][c-1] = -2*N

def rink(i):
  res = [0]*N
  for j in range(N):
    if R[i][j] > 1:
      res[j] = 1
  return res

def rev(ans,res):
  res = ans + res
  for i in range(N):
    if res[i] <= 0:
      res[i] = 0
    elif res[i] < N:
      res[i] = 1
    else:
      res[i] = 0
  return res

ANS = np.array([[0]*N]*N)
for p,r in enumerate(R):
  ans = r
  for i in range(N):
    if ans[i] > 0:
      res = rink(i)
      t = rev(ans,res)
      c = 0
      while sum(ans-t) != 0:
        if c > N:break
        c += 1
        ans = t
        for j in range(N):
          if t[j] > 0:
            res = rink(j)
            t = rev(t,res)
            
  ANS[p] = ans

def trim(x):
  if x == 1:
    return 1
  else:
    return 0


ANS = ANS + R
for i,a in enumerate(ANS):
  ANS[i] = list(map(trim, a))

ans = ''
for a in ANS:
  ans += str(sum(a)-1) + ' '

print(ans[:-1])