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
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines

class UnionFind_size:
    def __init__(self, n):
        # 初期状態は全要素が根なので、親は自分自身
        self.par = [i for i in range(n+1)]
        # 集団の要素数(size)を格納する(初期は全て1)
        self.size = [1] * (n+1)
    
    # 根を検索する関数
    def find(self, x):
        # 根までたどったらその番号を返す
        if self.par[x] == x:
            return x
        # 根でないなら親の要素で再検索
        else:
            # 検索過程で親を書き換える(圧縮経路)
            self.par[x] = self.find(self.par[x])
            return self.find(self.par[x])
  
    # 結合(unite)する関数
    def unite(self, x, y):
        # 根を探す
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        # サイズを比較し、小さい方を大きい方に繋げる
        if self.size[x] < self.size[y]:
            self.par[x] = y
            self.size[y] += self.size[x]
        else:
            self.par[y] = x
            self.size[x] += self.size[y]
           
    # 同じグループに属するかを判定する関数
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
    
    # 要素が属する木のサイズを返す関数
    def get_size(self, x):
        return self.size[self.find(x)]
    
    # グループ数を返す関数
    def group_sum(self):
        c = 0
        for i in range(len(self.par)):
            if self.find(i) == i:
                c += 1
        return c


def main():
    N,M,K = map(int, readline().split())

    uf = UnionFind_size(N+1)
    F = [[] for _ in range(N+1)]
    for _ in range(M):
        a,b = map(int, readline().split())
        F[a].append(b)
        F[b].append(a)
        uf.unite(a, b)
    
    B = [[] for _ in range(N+1)]
    for _ in range(K):
        c,d = map(int, readline().split())
        B[c].append(d)
        B[d].append(c)


    ans = []
    for i in range(1,N+1):
        tmp = 0
        tmp += uf.get_size(i) - 1
        tmp -= len(F[i])
        for j in B[i]:
            if uf.same_check(i, j):
                tmp -= 1
        ans.append(tmp)
    
    print(" ".join(map(str, ans)))


if __name__ == "__main__":
    main()