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)

class UnionFindWithRank(object):
    par = []
    rank = []
    cache = True
    log = []

    def __init__(self, n):
        for i in range(n):
            self.par.append(i)
            self.rank.append(0)

    def root(self, x):
        if self.par[x] == x:
            return x
        else:
            p = self.root(self.par[x])
            self.par[x] = p
            return p

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

    def unite(self, x, y):
        ox, oy = x, y
        x = self.root(x)
        y = self.root(y)

        if x == y:
            return

        if self.rank[x] < self.rank[y]:
            x, y = y, x
            self.par[y] = x
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

class UnionFind():
    # 作りたい要素数nで初期化
    # 使用するインスタンス変数の初期化
    def __init__(self, n):
        self.n = n
        # root[x]<0ならそのノードが根かつその値が木の要素数
        # rootノードでその木の要素数を記録する
        self.root = [-1] * (n + 1)
        # 木をくっつける時にアンバランスにならないように調整する
        self.rnk = [0] * (n + 1)

    # ノードxのrootノードを見つける
    def Find_Root(self, x):
        if (self.root[x] < 0):
            return x
        else:
            # ここで代入しておくことで、後の繰り返しを避ける
            self.root[x] = self.Find_Root(self.root[x])
            return self.root[x]

    # 木の併合、入力は併合したい各ノード
    def Unite(self, x, y):
        # 入力ノードのrootノードを見つける
        x = self.Find_Root(x)
        y = self.Find_Root(y)
        # すでに同じ木に属していた場合
        if (x == y):
            return
            # 違う木に属していた場合rnkを見てくっつける方を決める
        elif (self.rnk[x] > self.rnk[y]):
            self.root[x] += self.root[y]
            self.root[y] = x

        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            # rnkが同じ（深さに差がない場合）は1増やす
            if (self.rnk[x] == self.rnk[y]):
                self.rnk[y] += 1

    # xとyが同じグループに属するか判断
    def isSameGroup(self, x, y):
        return self.Find_Root(x) == self.Find_Root(y)

    # ノードxが属する木のサイズを返す
    def Count(self, x):
        return -self.root[self.Find_Root(x)]

n, m, k =map(int, input().split())

uf = UnionFind(n)
bd = dict()
fd = dict()

for i in range(m):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    fd[200000 * a + b] = True
    fd[200000 * b + a] = True
    uf.Unite(a, b)
for i in range(k):
    c,d = map(int, input().split())
    c -= 1
    d -= 1
    bd[200000 * c + d] = True
    bd[200000 * d + c] = True

res = [0] * n
for i in range(n):
    for j in range(i+1, n):
        if (200000 * i + j) not in fd and (200000 * i + j) not in bd:
            if uf.isSameGroup(i, j):
                res[i] += 1
                res[j] += 1
print(" ".join(list(map(str, res))))