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 Julia (0.5.0) (judge verdict: Time Limit Exceeded)

type UnionFind{T}
    parent:: Vector{T}  # parent[root] is the negative of the size

    function UnionFind{T}(nodes::T)
        if nodes <= 0
            throw(ArgumentError("invalid argument for nodes: $nodes"))
        end

        parent = -ones(T, nodes)
        new{T}(parent)
    end
end

UnionFind(nodes::Integer) = UnionFind{typeof(nodes)}(nodes)

function root{T}(uf::UnionFind{T}, x)::T
    if uf.parent[x] < 0
        return x
    else
        # uf.parent[x] = root{T}(uf, uf.parent[x])
        # return uf.parent[x]
        return uf.parent[x] = root(uf, uf.parent[x])
    end
end

function issame{T}(uf::UnionFind{T}, x::T, y::T)
    return root(uf, x) == root(uf, y)
end

function size{T}(uf::UnionFind{T}, x::T)::T
    return -uf.parent[root(uf, x)]
end

function unite!{T}(uf::UnionFind{T}, x::T, y::T)
    x = root(uf, x)
    y = root(uf, y)
    if x == y
        return false
    end
    if uf.parent[x] > uf.parent[y]
        x, y = y, x
    end
    # unite smaller tree(y) to bigger one(x)
    uf.parent[x] += uf.parent[y]
    uf.parent[y] = x
    return true
end

quit() = exit(0)
in_1() = parse(Int, readline())
in_a() = [parse(Int,s) for s in split(readline())]
in_t() = (in_a()...,)
yesno(b) = b ? "Yes" : "No"

function main()
  N, M, K = in_a()
  AB = [in_t() for _ in 1:M]
  CD = [in_t() for _ in 1:K]
  uf = UnionFind(N)
  for (a,b) in AB
    unite!(uf, a, b)
  end
  groupsize = zeros(Int, N)
  for i in 1:N
    groupsize[root(uf, i)] += 1
  end
  adjn = [count(identity, (a==i || b==i for (a,b) in AB)) for i in 1:N]
  map(1:N) do i
    ans = groupsize[root(uf, i)] - 1 - adjn[i]
    ans -= count(identity, (c==i && issame(uf, i, d) for (c, d) in CD))
    ans -= count(identity, (d==i && issame(uf, i, c) for (c, d) in CD))
  end |> arr->join(string.(arr), ' ') |> println
end

isempty(ARGS) && main()