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)

# Disjoint Set: Union Find Tree
class UnionFind:
    def __init__(self, N):
        self.N = N  # ノード数
        # 最初は全てが根であるとして初期化        
        self.parent = [-1] * N  # idxが各ノードのid, 値がそのノードの親ノードのid

    def find(self, x):  # ノードのidを受け取って一番上の親ノードのidを帰す
        if (self.parent[x] < 0):
            return x
        self.parent[x] = self.find(self.parent[x])  # 経由したノードすべての親を上書き（経路圧縮）
        return self.parent[x]

    def unite(self, x, y): # ノード番号を2つ受け取って、そのノード同士をつなげる処理を行う。
        # 引数のノードを直接つなぐ代わりに、親同士を連結する処理にする。
        rx = self.find(x) # xの根をrx
        ry = self.find(y) # yの根をry
        if (rx == ry): # xとyの根が同じ(=同じ木にある)時はそのまま
            return False

        if self.parent[rx] > self.parent[ry]:
            rx, ry = ry, rx
 
        self.parent[rx] += self.parent[ry]
        self.parent[ry] = rx

        return True
    
    def size(self, x):
        return -self.parent[self.find(x)]
      
    def same(self, x, y):
        return self.find(x) == self.find(y)
      
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
 
    def roots(self):
        return [i for i, x in enumerate(self.parent) if x < 0]
 
    def group_count(self):
        return len(self.roots())
 
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}
    
N, M, K = map(int, input().split())
ds = UnionFind(N+1)
num = [0]*(N+1)
G = [[] for i in range(N+1)]
for _ in range(M):
    x, y = map(int, input().split())
    G[x].append(y)
    G[y].append(x)
    ds.unite(x, y)
for _ in range(K):
    x, y = map(int, input().split())
    if ds.same(x,y) == True:
        G[x].append(y)
        G[y].append(x)
for i in range(1,N+1):
    print(ds.size(i) - len(G[i])-1,end=" ")