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)

def main():
    import sys
    buf = sys.stdin.buffer

    class UnionFind:
        def __init__(self, size):
            self.parent = [-1] * size
            self.rank = [1] * size

        def get_root(self, node):
            parent = self.parent[node]
            if parent == -1:
                root = node
            else:
                root = self.get_root(parent)
                self.parent[node] = root  # 同じnodeへの2回目以降のget_rootを高速にするために、直接rootに繋いでおく
            return root

        def in_same_group(self, node1, node2):
            root1 = self.get_root(node1)
            root2 = self.get_root(node2)
            return root1 == root2

        def unite(self, node1, node2):
            if self.in_same_group(node1, node2):
                return
            main_root = self.get_root(node1)
            sub_root = self.get_root(node2)
            if self.rank[main_root] < self.rank[sub_root]:  # rankの大きい方をmain_rootにする
                main_root, sub_root = sub_root, main_root

            self.parent[sub_root] = main_root
            self.rank[main_root] += self.rank[sub_root]

    n, m, k = map(int, buf.readline().split())
    uf = UnionFind(n)
    minus_count = [1] * n

    query = list(map(int, buf.read().split()))
    follows = query[:2*m]
    blocks = query[2*m:]

    for a, b in zip(follows[0::2], follows[1::2]):
        uf.unite(a-1, b-1)
        minus_count[a-1] += 1
        minus_count[b-1] += 1

    for a, b in zip(blocks[0::2], blocks[1::2]):
        if uf.in_same_group(a-1, b-1):
            minus_count[a-1] += 1
            minus_count[b-1] += 1

    result = [uf.rank[uf.get_root(i)] - minus_count[i] for i in range(n)]
    print(*result)


if __name__ == '__main__':
    main()