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 (3.8.2) (judge verdict: Accepted)

#!usr/bin/env pypy3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect


def LI(): return [int(x) for x in sys.stdin.readline().split()]


def LI1(): return [int(x) - 1 for x in sys.stdin.readline().split()]


def I(): return int(sys.stdin.readline())


def LS(): return [list(x) for x in sys.stdin.readline().split()]


def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res


def IR(n):
    return [I() for i in range(n)]


def LIR(n):
    return [LI() for i in range(n)]


def SR(n):
    return [S() for i in range(n)]


def LSR(n):
    return [LS() for i in range(n)]


sys.setrecursionlimit(1000000)


class UnionFind:
    def __init__(self, n):
        self._parent = [i for i in range(n)]
        self._rank = [0 for _ in range(n)]
        self._group_size = [1 for _ in range(n)]
        self.num_of_groups = n

    def find(self, x):
        if self._parent[x] == x:
            return x
        px = self._parent[x]
        root = self.find(px)
        self._parent[x] = root
        return root

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if px == py:
            return
        if self._rank[px] < self._rank[py]:
            self._parent[px] = py
            self._group_size[py] += self._group_size[px]
        else:
            self._parent[py] = px
            self._group_size[px] += self._group_size[py]
        if self._rank[px] == self._rank[py]:
            self._rank[py] += 1
        self.num_of_groups -= 1

    def is_same(self, x, y):
        return self.find(x) == self.find(y)

    def group_size(self, x):
        px = self.find(x)
        return self._group_size[px]


def main():
    N, M, K = LI()
    AB = [LI1() for _ in range(M)]
    CD = [LI1() for _ in range(K)]

    uf = UnionFind(N)

    friend = [[] for _ in range(N)]
    block = [[] for _ in range(N)]

    for (a, b) in AB:
        friend[a].append(b)
        friend[b].append(a)
        uf.union(a, b)
    for (c, d) in CD:
        block[c].append(d)
        block[d].append(c)

    for i in range(N):
        gs = uf.group_size(i)
        bc = 0
        for bi in block[i]:
            bc += int(uf.is_same(i, bi))
        # print(i, gs, fc, bc)
        print(gs - len(friend[i]) - bc - 1, end=" ")


main()