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: Time Limit Exceeded)

# xの根を求める
def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]


# xとyの属する集合の併合
def unite(x, y):
    x = find(x)
    y = find(y)

    if x == y:
        return False
    else:
        # sizeの大きいほうがx
        if par[x] > par[y]:
            x, y = y, x
        par[x] += par[y]
        par[y] = x
        return True


# xとyが同じ集合に属するかの判定
def same(x, y):
    return find(x) == find(y)


# xが属する集合の個数
def size(x):
    return -par[find(x)]


n, m, k = map(int, input().split())
ab = [list(map(int, input().split())) for i in range(m)]
cd = [list(map(int, input().split())) for i in range(k)]

# 初期化
# 根なら-size,子なら親の頂点
par = [-1] * n

friend = [[0] * n for i in range(n)]
block = [[0] * n for i in range(n)]

not_see = [set() for i in range(n)]

for a, b in ab:
    a -= 1
    b -= 1
    unite(a, b)
    friend[a][b] = 1
    friend[b][a] = 1
    if same(a, b):
        not_see[a].add(b)
        not_see[b].add(a)

for c, d in cd:
    c -= 1
    d -= 1
    block[c][d] = 1
    block[d][c] = 1
    if same(c, d):
        not_see[c].add(d)
        not_see[d].add(c)

for i in range(n):
    print(size(i) - len(not_see[i]) - 1)