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)

n, m, k = map(int, input().split())
a_m, b_m = [], []
c_k, d_k = [], []
def minus(x):
    return int(x)-1

friends_list = [[] for i in range(n)]
block_list = [[] for i in range(n)]
for i in range(m):
    a_m.append(0)
    b_m.append(0)
    a_m[i], b_m[i] = map(minus, input().split())
    friends_list[a_m[i]].append(b_m[i])
    friends_list[b_m[i]].append(a_m[i])
for i in range(k):
    c_k.append(0)
    d_k.append(0)
    c_k[i], d_k[i] = map(minus, input().split())
    block_list[c_k[i]].append(d_k[i])
    block_list[d_k[i]].append(c_k[i])

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents =[-1 for i in range(n)]
    # どのグループに属しているかを返す
    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 unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return False
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        self.parents[x] += self.parents[y]
        self.parents[y] = x
        return True

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

def blocks_in_union(x):
    count = 0
    for block in block_list[x]:
        if uf.find(x) == uf.find(block):
            count += 1
    return count

uf = UnionFind(n)
for i in range(m):
    uf.unite(a_m[i], b_m[i])
for i in range(n):
    print(uf.size(i)-1-len(friends_list[i])-blocks_in_union(i), end=' ')