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 (3.8.2) (judge verdict: Wrong Answer)

import sys
import copy as cp
from collections import defaultdict as dd
sys.setrecursionlimit(10**8)

def find(x):
    if par[x]==x:
        return x
    else:
        par[x]=find(par[x])
        return par[x]

def union(a,b):
    a=find(a)
    b=find(b)
    if a==b:
        return
    
    if rank[a]<rank[b]:
        par[a]=b
        rank[b]+=rank[a]
    else:
        par[b]=a
        rank[a]+=rank[b]       

    return
 

N,M,K=map(int, input().split())

par=(list(range(N+1)))
rank=[1]*(N+1)

path=dd(list)  ##直接つながっている枝

for _ in range(M):
    A,B=map(int, input().split())
    path[A].append(B)
    path[B].append(A)
 
#print('path')
#print(path)

fd=dd(list)   ##ブロック関係
for _ in range(K):
    C,D=map(int, input().split())
    fd[C].append(D)  ##ブロック関係をメモ
    fd[D].append(C)

#print('fd')
#print(fd)
      
for i in path:
    for j in path[i]:
        union(i,j)

fl=dd(list)  ##フレンド候補
ans=[]
for i in range(1,N+1):
    for j in range(i+1,N+1):
        if par[i]==par[j]:
            if j not in fd[i] and j not in path[i]:
                fl[i].append(j)
            if i not in fd[j] and i not in path[j]:
                fl[j].append(i)  
    ans.append(len(fl[i]))

print(' '.join(map(str,ans)))