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)

#!/usr/bin/env python3
import sys
from collections import defaultdict

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
        self.ranks = [0] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return
        
        if self.ranks[x] < self.ranks[y]:
            self.parents[x] = y
        else:
            self.parents[y] = x
            if self.ranks[x] == self.ranks[y]:
                self.ranks[x]+=1

        # if self.parents[x] > self.parents[y]:
        #     x, y = y, x

        # self.parents[x] += self.parents[y]
        # self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def tup(a,b):
    return (min(a,b),max(a,b))

def solve(N: int, M: int, K: int, A: "List[int]", B: "List[int]", C: "List[int]", D: "List[int]"):
    friends = defaultdict(bool)
    group = UnionFind(N)
    for i in range(M):
        friends[tup(A[i]-1,B[i]-1)] = True
        group.union(A[i]-1,B[i]-1)
    for i in range(K):
        friends[tup(C[i]-1,D[i]-1)] = True
    result = ""
    isFirst = True
    roots = group.roots()
    rootSets = {r: set() for r in roots }
    for i in range(N):
        r = group.find(i)
        rootSets[r].add(i)
    # for i in range(N):
    #     c = 0
    #     r = group.find(i)
    #     for g in rootSets[r]:
    #         if g != i and not friends[tup(i,g)]:
    #             c+= 1
    #     if not isFirst:
    #         result+=" "
    #     else:
    #         isFirst = False
    #     result+=str(c)
    print(result)
    return


# Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    K = int(next(tokens))  # type: int
    A = [int()] * (M)  # type: "List[int]"
    B = [int()] * (M)  # type: "List[int]"
    for i in range(M):
        A[i] = int(next(tokens))
        B[i] = int(next(tokens))
    C = [int()] * (K)  # type: "List[int]"
    D = [int()] * (K)  # type: "List[int]"
    for i in range(K):
        C[i] = int(next(tokens))
        D[i] = int(next(tokens))
    solve(N, M, K, A, B, C, D)

if __name__ == '__main__':
    main()