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)

import sys
sys.setrecursionlimit(1000000000)
import math
from fractions import gcd
def lcm(a, b): return a * b // gcd(a, b)
from itertools import count, permutations
from functools import lru_cache
from collections import deque, defaultdict
from pprint import pprint
ii = lambda: int(input())
mis = lambda: map(int, input().split())
lmis = lambda: list(mis())
INF = float('inf')
N1097 = 10**9 + 7

def meg(f, ok, ng):
    while abs(ok-ng)>1:
        mid = (ok+ng)//2
        if f(mid):
            ok=mid
        else:
            ng=mid
    return ok

def get_inv(n, modp):
    return pow(n, modp-2, modp)

def factorials_list(n, modp):    # 10**6
    fs = [1]
    for i in range(1, n+1):
        fs.append(fs[-1] * i % modp)
    return fs

def invs_list(n, fs, modp):     # 10**6
    invs = [get_inv(fs[-1], modp)]
    for i in range(n, 1-1, -1):
        invs.append(invs[-1] * i % modp)
    invs.reverse()
    return invs

def comb(n, k, modp):
    num = 1
    for i in range(n, n-k, -1):
        num = num * i % modp
    den = 1
    for i in range(2, k+1):
        den = den * i % modp
    return num * get_inv(den, modp) % modp

def comb_from_list(n, k, modp, fs, invs):   
    return fs[n] * invs[n-k] * invs[k] % modp

#

class UnionFindEx:
    def __init__(self, size):
        self.roots = [{i} for i in range(size)]
    def getRootID(self, i):
        r = self.roots[i]
        if isinstance(r, set):
            return i
        else:
            r = self.getRootID(r)
            self.roots[i] = r
            return r
    def getGroupSize(self, i):
        return len(self.getGroupSet(i))
    def getGroupSet(self, i):
        return self.roots[self.getRootID(i)]
    def connect(self, i, j):
        r1, r2 = self.getRootID(i), self.getRootID(j)
        if r1 == r2:
            return False
        if self.getGroupSize(r1) < self.getGroupSize(r2):
            r1, r2 = r2, r1
        self.roots[r1] |= self.roots[r2]
        self.roots[r2] = r1
        return True

def main():
    N,M,K=mis()
    uf = UnionFindEx(N)
    frs = tuple(set() for _ in range(N))
    bls = tuple(set() for _ in range(N))
    for _ in range(M):
        a, b = mis()
        uf.connect(a-1, b-1)
        frs[a-1].add(b-1)
        frs[b-1].add(a-1)
    for _ in range(K):
        a, b = mis()
        bls[a-1].add(b-1)
        bls[b-1].add(a-1)
    print(*(len(uf.getGroupSet(i) - frs[i] - bls[i]) - 1 for i in range(N)))


main()