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: Accepted)

#import sys
#sys.setrecursionlimit(500000)　マージテクをちゃんとしていればこんなものはいらないはず……

class UF:
    #経路圧縮＋マージテクのunion find木
    def __init__(self,n):
        self.n = n
        self.parents = [-1] * (n+1) #1-indexedにするために１個足している
        self.rnks = [0] * (n+1)

    def find(self,n):
        if self.parents[n] < 0:
            return n
        else:
            self.parents[n] = self.find(self.parents[n])
            return self.parents[n]

    def union(self,a,b):
        a_par = self.find(a)
        b_par = self.find(b)

        if a_par == b_par:
            return
        if self.rnks[a_par] < self.rnks[b_par]: #ランクの高い方の根に低い方の根をつける　逆にしないように……！
            a_par,b_par = b_par,a_par
        if self.rnks[a_par] == self.rnks[b_par]:
            self.rnks[a_par] += 1
        self.parents[a_par] += self.parents[b_par]
        self.parents[b_par] = a_par

    def length(self,n):
        par = self.find(n)
        return abs(self.parents[par])

    def is_same(self,a,b):
        return self.find(a) == self.find(b)



def main():
    n,m,k = map(int,input().split())
    uf = UF(n)
    block_lst = [0 for _ in range(n+1)]
    friend_lst = [0 for _ in range(n+1)]
    #友達リスト部分
    for i in range(m):
        a,b = map(int,input().split())
        friend_lst[a] += 1
        friend_lst[b] += 1
        uf.union(a,b)
    #print(friend_lst)

    #ブロックリスト部分
    for i in range(k):
        c,d = map(int,input().split())
        if uf.is_same(c,d):#c,dが同じ友達候補グループにない限り、解答には影響しない！
            block_lst[c] += 1
            block_lst[d] += 1
    #print(block_lst)
    #メイン処理
    answer = []

    for i in range(1,n+1):
        f_o_f = uf.length(i)
        answer.append(f_o_f - friend_lst[i] - block_lst[i] - 1)


    print(" ".join(map(str,answer)))






if __name__ == '__main__':
    main()