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: Accepted)

import sys
import collections

class UnionFind:
    def __init__(self, size):
        self.parent = [-1] * (size + 1)  # 非負なら親ノード，負ならグループの要素数

    def root(self, x):  # root(x): xの根ノードを返す．
        if self.parent[x] < 0:
            return x
        elif self.parent[self.parent[x]] < 0:
            return self.parent[x]
        else:
            self.parent[x] = self.root(self.parent[x])  # xをxの根に直接つなぐ
            return self.parent[x]

    def merge(self, x, y):  # merge(x,y): xのいるグループと$y$のいるグループをまとめる
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return False
        if self.parent[x] > self.parent[y]:  # xの要素数がyの要素数より「小さい」とき入れ替える
            x, y = y, x
        self.parent[x] += self.parent[y]  # xの要素数を更新
        self.parent[y] = x  # yをxにつなぐ
        return True

    def issame(self, x, y):  # same(x,y): xとyが同じグループにあるならTrue
        return self.root(x) == self.root(y)

    def size(self, x):  # size(x): xのいるグループの要素数を返す
        return -self.parent[self.root(x)]


def do():
    sys.setrecursionlimit(100000000)

    N, M, K = map(int, input().split())
    friends = [list(map(int, input().split())) for _ in range(M)]
    blocks = [list(map(int, input().split())) for _ in range(K)]

    uf1 = UnionFind(N + 1) # 友達候補のUF
    friends_list= [set() for _ in range(N + 1)]
    block_list = [set() for _ in range(N + 1)]

    for query in friends :
        a,b= query
        uf1.merge(a,b)
        friends_list[a].add(b)
        friends_list[b].add(a)

    for query in blocks:
        a, b = query
        block_list[a].add(b)
        block_list[b].add(a)
    anss = []
    for i in range(1,N+1):
        s = uf1.size(i) -1 #ベースの人数

        friends = friends_list[i]
        for friend in friends:
            if uf1.issame(friend,i):
                s -=1

        blocks = block_list[i]
        for block in blocks:
            if uf1.issame(block,i):
                s -=1
        anss.append(s)
    print(*(anss))

if __name__ == "__main__":
    do()