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: Accepted)

parseInt(x) = parse(Int, x)
parseBigInt(x) = parse(BigInt, x)
parseFloat(x) = parse(Float64, x)
parseBigFloat(x) = parse(BigFloat, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function root(par::Vector{Int},x::Int)
    index=par[x]
    while par[index]!=index
        index=par[index]
    end
    return index
end

function unite(par::Vector{Int},siz::Vector{Int},x::Int,y::Int)
    rx,ry=root(par,x),root(par,y)
    if rx==ry
        return 0
    elseif rx>ry
        rx,ry=ry,rx
    end
    par[ry]=rx
    siz[rx]+=siz[ry]
end

function same(par::Vector{Int},x::Int,y::Int)
    return root(par,x)==root(par,y)
end

function main()
    n,m,k=parseMap(split(readline()))
    par=[i for i in 1:n]
    siz=[1 for i in 1:n]
    ans=[-1 for i in 1:n]   # 自分を除いておく

    # 友達候補をUnion-Findで実装
    for i in 1:m
        a,b=parseMap(split(readline()))
        unite(par,siz,a,b)
        ans[a]-=1   # 友達なら候補ではないので-1
        ans[b]-=1   # 友達なら候補ではないので-1
    end

    # ブロック関係の相手を友達関係から除外
    for i in 1:k
        c,d=parseMap(split(readline()))
        if same(par,c,d)
            ans[c]-=1   # 友達候補に含まれていれば除外
            ans[d]-=1   # 友達候補に含まれていれば除外
        end
    end

    for i in 1:n
        ans[i]+=siz[root(par,i)]
    end
    println(join(ans," "))
end
main()