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 (3.8.2) (judge verdict: Accepted)

class UnionFind():
    def __init__(self, n):
        self._n = n
        self._table = [-1]*n

    def _root(self, x):
        stack = []
        while self._table[x] >= 0:
            stack.append(x)
            x = self._table[x]
        for y in stack:
            self._table[y] = x
        return x
    
    def unite(self, x, y):
        x, y = self._root(x), self._root(y)
        if x == y:
            return
        if x > y:
            x, y = y, x
        self._table[x] += self._table[y]
        self._table[y] = x
    
    def same(self, x, y):
        return self._root(x) == self._root(y)

    def count_members(self, x):
        return -self._table[self._root(x)]
    
    def count_groups(self):
        return len({self._root(i) for i in range(n)})
    
    def __str__(self):
        return str([self._root(i) for i in range(n)])
    
    def __repr__(self):           
        return repr([self._root(i) for i in range(n)])

n, m, k, *ABCD = map(int, open(0).read().split())
AB = zip(ABCD[:2*m][::2], ABCD[:2*m][1::2])
CD = zip(ABCD[2*m:][::2], ABCD[2*m:][1::2])

uf = UnionFind(n)

friends = [set() for _ in range(n)]
for a, b in AB:
    a, b = a-1, b-1
    uf.unite(a, b)
    friends[a].add(b)
    friends[b].add(a)
    
blocks = [set() for _ in range(n)]
for c, d in CD:
    c, d = c-1, d-1
    if uf.same(c, d):
        blocks[c].add(d)
        blocks[d].add(c)

candidates = []
for i in range(n):
    candidates.append(uf.count_members(i)-len(friends[i])-len(blocks[i])-1)
print(*candidates)