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)

from collections import deque
import sys
sys.setrecursionlimit(4100000)

n, m, k = map(int, input().split())

#友達関係
F = [deque([]) for i in range(n)] #Friend
for i in range(m):
    a, b = map(int, input().split())
    F[a-1].append(b)
    F[b-1].append(a)

#DFSで連結部分ごとのグループに分ける
G = [0] * n #Group
GS = deque([])   #Group Size
gid = 0 #group id

#DFSを行うための関数
def using_DFS(p):
    if G[p] == 0:
        G[p] = gid
        for j in F[p]:
            if G[j-1] == 0:
                using_DFS(j-1)
            else:   #すでに登場
                pass
    else:
        pass

for i in range(n):
    if G[i] == 0:
        gid += 1
        t = 0
        using_DFS(i)
        GS.append(G.count(gid))
    else:
        pass
#print('G=',G)
#print('S=',S)

#ブロック関係
BinG = [deque([]) for i in range(n)] #Block in Group
for i in range(k):
    c, d = map(int, input().split())
    if G[c-1] == G[d-1]:
        BinG[c-1].append(d)
        BinG[d-1].append(c)
    else:
        pass
#print('F=',F)
#print('B=',B)

for i in range(n):
    ans = GS[G[i]-1] - 1 - len(F[i]) - len(BinG[i])
    print(ans)