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)

import sys
from math import *
from collections import deque, Counter, defaultdict
from fractions import gcd
from copy import deepcopy
input = lambda: sys.stdin.readline().rstrip()

def eprint(s):
    sys.stderr.write('DEBUG: {}'.format(s))
    return

def main():
    n, m, k = map(int, input().split())
    fgraph = [list() for i in range(n)]
    bgraph = [list() for i in range(n)]
    cgraph = [list() for i in range(n)]
    
    is_all_seen = [False]*n
    cnt = 0
    sn = [None]*n
    rks = [None] *n
    sys.setrecursionlimit(10**9)
    def dfs(graph,s,r):
        seen[s] = s+1
        if rks[s] is None:
            rks[s] = r

        for i in range(len(graph[s])):
            if seen[graph[s][i]-1]:
                continue
            dfs(graph, graph[s][i]-1,r)

    for i in range(m):
        a,b = map(int, input().split())

        fgraph[a-1].append(b)
        fgraph[b-1].append(a)

    # print(fgraph)
    #print(bgraph)
    # 連結成分に対してid的なindexをつける
    c = -1
    for i in range(n):
        seen = deepcopy(sn)
        if rks[i] is None:
            c += 1
            dfs(fgraph, i, c)
    
    for i in range(k):
        c,d = map(int, input().split())
        if rks[c-1] == rks[d-1]:
            bgraph[d-1].append(c)
            bgraph[c-1].append(d)

    counter = Counter(rks)
    for i in range(n):
        # counter[rks[i]] 連結成分のサイズ
        # len(fgraph[i]) 友達の数
        # len(bgraph[i]) 連結成分内のブロックの数
        # 1 自分
        print(counter[rks[i]] - len(fgraph[i]) - len(bgraph[i]) - 1)

if __name__ == '__main__':
    main()