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: Accepted)

class UnionFind:

    def __init__(self, n):
        self.p, self.d = [-1] * n, [0] * n

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

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

    def unite(self, x, y):
        if self.isSame(x, y):
            return
        x, y = self.find(x), self.find(y)

        if self.d[x] > self.d[y]:
            self.p[x] += self.p[y]
            self.p[y] = x
        else:
            self.p[y] += self.p[x]
            self.p[x] = y
            if self.d[x] == self.d[y]:
                self.d[y] += 1

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


N, M, K = map(int, input().split())
n, u = {i: set() for i in range(N)}, UnionFind(N)
for _ in range(M):
    A, B = map(int, input().split())
    u.unite(A - 1, B - 1)
    n[A - 1].add(B - 1)
    n[B - 1].add(A - 1)
for _ in range(K):
    C, D = map(int, input().split())
    if u.isSame(C - 1, D - 1):
        n[C - 1].add(D - 1)
        n[D - 1].add(C - 1)

print(' '.join([str(u.size(i) - len(n[i]) - 1) for i in range(N)]))