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)

# -*- coding: utf-8 -*-

def display(node):
        print('----')
        print('val: {}'.format(node.val))
        print('friend_numbers: {}'.format(node.friend_numbers))
        print('block_numbers: {}'.format(node.block_numbers))
        print('candidate_numbers: {}'.format(node.candidate_numbers))

def find_friend(root, target):
    #queue = [root]
    stack = [root]
    already_list = [root.val]
    while len(stack) > 0:
        node = stack.pop(-1)
        if target in node.friend_numbers:
            return True
        if target in node.candidate_numbers:
            return True
        
        for friend in node.friends:
            if friend.val not in already_list:
                stack.append(friend)
                already_list.append(friend.val)

#def find_friend(tree_node, target, already_list):
#    for friend in tree_node.friends:
#        if target in friend.friend_numbers:
#            return True
#        if target in friend.candidate_numbers:
#            return True
#
#    flags = []
#    for friend in tree_node.friends:
#        if friend.val not in already_list:
#            new_already_list = already_list[:]
#            new_already_list.append(friend.val)
#            flags.append(find_friend(friend, target, new_already_list))
#    if True in flags:
#        return True
#    return False


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.friends = []
        self.friend_numbers = []
        self.blocks = []
        self.block_numbers = []
        self.candidate_numbers = []

[N, M, K] = [int(i) for i in input().split()]

AB = []
for m in range(M):
    [a, b] = [int(i) for i in input().split()]
    AB.append((a, b))

CD = []
for k in range(K):
    [c, d] = [int(i) for i in input().split()]
    CD.append((c, d))

tree_nodes = {}
for i in range(1, N + 1):
    tree_node = TreeNode(i)
    tree_nodes[i] = tree_node

for (a, b) in AB:
    tree_nodes[a].friends.append(tree_nodes[b])
    tree_nodes[a].friend_numbers.append(b)
    tree_nodes[b].friends.append(tree_nodes[a])
    tree_nodes[b].friend_numbers.append(a)

for (c, d) in CD:
    tree_nodes[c].blocks.append(tree_nodes[d])
    tree_nodes[c].block_numbers.append(d)
    tree_nodes[d].blocks.append(tree_nodes[c])
    tree_nodes[d].block_numbers.append(c)

candidates = []
for k, v in tree_nodes.items():
    # 友達候補を探す
    candidate = 0
    for n in range(1, N + 1):
        if n == k:
            continue
        if n in v.friend_numbers:
            continue
        if n in v.block_numbers:
            continue

        flag = False
        # nさんが既に自分を友達候補にしていればOK
        if k in tree_nodes[n].candidate_numbers:
            flag = True
        else:
            # nさんへの友達経路があるか探索
            #flag = find_friend(v, n, [v.val])
            flag = find_friend(v, n)
        if flag:
            candidate += 1
            v.candidate_numbers.append(n)
        #print('{}の友達候補{}: {}'.format(k, n, flag))
    
    candidates.append(candidate)

ans = ''
for candidate in candidates:
    ans += str(candidate)
    ans += ' '

#for k, v in tree_nodes.items():
#    display(v)

print(ans[:len(ans) - 1])