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 (7.3.0) (judge verdict: Accepted)

import sys
from collections import deque

sys.setrecursionlimit(10**7)
input = sys.stdin.readline

n,m,k = map(int, input().split())
f = [[] for _ in range(n)]
g = [[] for _ in range(n)]

for _ in range(m):
    a, b = map(int, input().split())
    a, b = a - 1, b - 1
    f[a].append(b)
    f[b].append(a)

for _ in range(k):
    c, d = map(int, input().split())
    c, d = c - 1, d - 1
    g[c].append(d)
    g[d].append(c)

d = {}
parent = [-1] * n
seen = [False] * n

# DFS
for i in range(n):
    if seen[i]:
        continue
    stack = deque()
    # グループは集合で管理
    d[i] = set([i])
    stack = [i]
    while stack:
        v = stack.pop()
        seen[v] = True
        # 親を設定して同じ連結成分であることを判別する
        parent[v] = i
        for nv in f[v]:
            if seen[nv]:
                continue
            # 集合に追加する
            d[i].add(nv)
            stack.append(nv)

ans = [0] * n
for i in range(n):
    # 連結成分ごとに集合を取り出す
    group = d[parent[i]]
    # 集合の全要素数 - 友達 - ブロック - 自分
    sum = len(group) - len(f[i]) - len(set(g[i]) & group) - 1
    ans[i] = sum

print(*ans)