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: Time Limit Exceeded)

module S = Set.Make (struct type t = int let compare = compare end) ;;

Scanf.scanf "%d %d %d" (fun n m k ->
    let friend = Array.make n [] in
    let block = Array.make n S.empty in
    for i = 1 to m do
        Scanf.scanf " %d %d" (fun a b ->
            let a = a - 1 in let b = b - 1 in
            friend.(a) <- b :: friend.(a);
            friend.(b) <- a :: friend.(b);
        )
    done;
    for i = 1 to k do
        Scanf.scanf " %d %d" (fun c d ->
            let c = c - 1 in let d = d - 1 in
            block.(c) <- S.add d block.(c);
            block.(d) <- S.add c block.(d);
        )
    done;
    let clust = Array.make n (-1) in
    let cm    = Array.make n [||] in
    let rec loop i cl =
        if i < n then (
            let cl = if clust.(i) <> -1 then cl else (
                let rec collect acc next = function
                    | [] -> if next = [] then acc else (collect acc [] next)
                    | hd :: tl ->
                            let acc = hd :: acc in
                            let next = List.fold_left (fun next v -> if clust.(v) <> -1 then next else (clust.(v) <- cl; v :: next)) next friend.(hd) in
                            collect acc next tl
                in
                clust.(i) <- cl;
                let acc = collect [] [i] [] in
                cm.(cl) <- Array.of_list acc;
                cl + 1
            )
            in
            let b = Array.fold_left (fun acc v -> if S.mem v block.(i) then acc + 1 else acc) 0 cm.(clust.(i)) in

            (* let c = S.cardinal (S.diff (S.diff cm.(clust.(i)) block.(i)) friend.(i)) in *)
            let c = Array.length cm.(clust.(i)) - List.length (friend.(i)) - b in

            let () = Printf.printf "%d " (c - 1) in
            loop (i + 1) cl
        )
    in
    loop 0 0
)