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 Ruby (2.7.1) (judge verdict: Accepted)

N,M,K = gets.split.map(&:to_i)

友達の輪と大きさ = [-1]*(N+1)
輪 = lambda{|a|
	next 友達の輪と大きさ[a]<0 ? a : 友達の輪と大きさ[a]=輪[友達の輪と大きさ[a]]
}
輪の大きさ = lambda{|a|
	next -友達の輪と大きさ[輪[a]]
}
友達です = lambda{|*ab|
	a,b = ab.map(&輪)
	next if a==b
	a,b = b,a if 友達の輪と大きさ[b]<友達の輪と大きさ[a]
	友達の輪と大きさ[a] += 友達の輪と大きさ[b]
	友達の輪と大きさ[b] = a
}

友達の数 = [0]*(N+1)

M.times{
	a,b = gets.split.map(&:to_i)
	友達です[a,b]
	友達の数[a] += 1
	友達の数[b] += 1
}

友達の輪の中の目障りなあいつ = [0]*(N+1)

K.times{
	a,b = gets.split.map(&:to_i)
	next if 輪[a]!=輪[b]
	友達の輪の中の目障りなあいつ[a] += 1
	友達の輪の中の目障りなあいつ[b] += 1
}

puts 1.upto(N).map{|n|
	輪の大きさ[n] - 1 - 友達の数[n] - 友達の輪の中の目障りなあいつ[n]
}