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)

class UnionFind():
    # コンストラクタ
    def __init__(self, n):
        # 要素数
        self.n = n
        # parents[i]: 要素iの親要素の番号
        # 要素iが根の場合、parents[i] = -(そのグループの要素数)
        self.parents = [-1] * n
 
    # 要素xが属するグループの根を返す
    # 経路圧縮
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
 
    # 要素xが属するグループと要素yが属するグループとを併合する
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
 
        if x == y:
            return
 
        if self.parents[x] > self.parents[y]:
            x, y = y, x
 
        self.parents[x] += self.parents[y]
        self.parents[y] = x
 
    # 要素xが属するグループの要素数を返す
    def size(self, x):
        return -self.parents[self.find(x)]
 
    # 要素x,yが同じグループに属するかどうかを返す
    def same(self, x, y):
        return self.find(x) == self.find(y)
 
    # 要素xが属するグループに属する要素をリストで返す
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
 
    # 全ての根の要素をリストで返す
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
 
    # グループの数を返す
    def group_count(self):
        return len(self.roots())

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

    best_friend = [0] * N
    friend = UnionFind(N)
    for _ in range(M):
        A,B = map(int,input().split())
        best_friend[A-1] += 1
        best_friend[B-1] += 1
        friend.union(A-1, B-1)

    blocked = [[] for _ in range(N)]
    for _ in range(K):
        C,D = map(int,input().split())
        blocked[C-1].append(D-1)
        blocked[D-1].append(C-1)

    ans = [0] * N
    for i in range(N):
        # 友達候補 - 友達 - 自分
        ans[i] = friend.size(i) - best_friend[i] - 1
        # 嫌いな人は友達になれない
        for bad in blocked[i]:
            if friend.same(i, bad):
                ans[i] -= 1

    print(*ans, sep=" ")

if __name__ == "__main__":
    main()