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 OCaml (4.02.3) (judge verdict: Accepted)

Scanf.scanf "%d %d %d" (fun n m k ->
    let par = Array.init n (fun i -> i) in
    let size = Array.make n 1 in
    let near = Array.make n 0 in
    let rec root x =
        if par.(x) = x then x else
            let r = root par.(x) in
            let () = par.(x) <- r in
            r
    in
    for i = 1 to m do
        Scanf.scanf " %d %d" (fun a b ->
            let a = a - 1 in let b = b - 1 in
            near.(a) <- near.(a) + 1;
            near.(b) <- near.(b) + 1;
            if root a <> root b then (
                size.(par.(a)) <- size.(par.(a)) + size.(par.(b));
                par.(par.(b)) <- par.(a)
            )
        )
    done;
    for i = 1 to k do
        Scanf.scanf " %d %d" (fun c d ->
            let c = c - 1 in let d = d - 1 in
            if root c = root d then (
                near.(c) <- near.(c) + 1;
                near.(d) <- near.(d) + 1;
            )
        )
    done;
    for i = 0 to n - 1 do
        Printf.printf "%d " (size.(root i) - near.(i) - 1)
    done
)