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:
    def __init__(self, num):
        self.rank = [0] * num
        self.par = [i for i in range(num)]

    def find_root(self, node):
        if self.par[node] == node:
            return node
        else:
            self.par[node] = self.find_root(self.par[node])
            return self.par[node]

    def same_root(self, x, y):
        return self.find_root(x) == self.find_root(y)

    def union(self, x, y):
        x = self.find_root(x)
        y = self.find_root(y)
        if x == y:
            return
        if self.rank[x] > self.rank[y]:
            self.par[y] = x
        else:
            self.par[x] = y
            if self.rank[x] == self.rank[y]:
                self.rank[y] += 1


# Input initial information.
n, m, k = map(int, input().split())
union_find = UnionFind(n)

# Input friend information and union.
direct_friend = [0] * n
for i in range(m):
    a, b = map(int, input().split())
    union_find.union(a - 1, b - 1)
    direct_friend[a - 1] += 1
    direct_friend[b - 1] += 1

# Group connected people.
group = [[] for _ in range(n)]
for i in range(n):
    group[union_find.find_root(i)].append(i)

# Get the count of the group for each (excluding myself).
res = [0] * n
for i in range(n):
    for people in group[i]:
        res[people] = len(group[i]) - 1

# Subtract the number of direct_friend from result.
for i in range(n):
    res[i] -= direct_friend[i]

# Input block information and subtract from result if both are in the same group.
for i in range(k):
    c, d = map(int, input().split())
    if union_find.same_root(c - 1, d - 1):
        res[c - 1] -= 1
        res[d - 1] -= 1

# Print the result.
print(*res)