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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
from collections import defaultdict

sys.setrecursionlimit(2147483647)


class Graph:
    def __init__(self):
        self.graph = defaultdict(set)

    def interconnect(self, u, v):
        self.graph[u].add(v)
        self.graph[v].add(u)

    def dfs(self, start):
        seen = set()
        return self.dfs_recurse(start, seen)

    def dfs_recurse(self, v, seen):
        seen.add(v)
        for u in self.graph[v] - seen:
            self.dfs_recurse(u, seen)
        return seen


if __name__ == "__main__":
    n, m, k = map(int, input().split())

    friends = Graph()
    blocked = defaultdict(set)

    [friends.interconnect(*map(lambda x: int(x) - 1, input().split())) for _ in range(m)]
    for _ in range(k):
        c, d = map(lambda x: int(x) - 1, input().split())
        blocked[c - 1].add(d)
        blocked[d - 1].add(c)

    visited = [False] * n
    counts = [0] * n

    for v in range(n):
        if visited[v]:
            continue
        elems = friends.dfs(v)
        elems_cnt = len(elems)
        for i in elems:
            counts[i] = elems_cnt - len(elems & friends.graph[i]) - len(elems & blocked[i]) - 1
            visited[i] = True

    print(*counts)