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)

from collections import deque

n, m, k = map(int, input().split())
fr = [list(map(int, input().split())) for i in range(m)]
br = [list(map(int, input().split())) for i in range(k)]

tb = [[None]*n for i in range(n)]
# 友達登録
for i in range(m):
    tb[fr[i][0]-1][fr[i][1]-1] = 1
    tb[fr[i][1]-1][fr[i][0]-1] = 1
# ブロック登録
for i in range(k):
    tb[br[i][0]-1][br[i][1]-1] = -1
    tb[br[i][1]-1][br[i][0]-1] = -1
# 自分は無視
for i in range(n):
    tb[i][i] = 0

#print("- - - - - - -")
#print(*tb, sep="\n")

def dfs_check():
    ans = [0 for i in range(n)]

    for i in range(n):
        #print("-- ",i+1,"の友達候補探し")
        #表の初期化
        g = [None for i in range(n)]

        stack = deque()
        stack.append(i)
        #最初の地点のコストは0
        g[i] = 0

        while stack:
            k = stack.pop() #stackから取り出す
            move = [d for d in range(n) if tb[k][d] == 1]#友達なら移動
            moveprint = [d+1 for d in range(n) if tb[k][d] == 1]
            #print(k+1, "の友達", moveprint)
            for dk in move:
                if g[dk] == None:
                    g[dk] = 1
                    stack.append(dk)
                    if tb[i][dk] == None:
                        #print(i+1,dk+1,"は友達！")
                        ans[i] += 1
                        ans[dk] += 1

    return ans

#友達を辿ってたどり着けるところまで考える
print(*dfs_check(), sep=" ")