Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and b_i.
Additionally, each vertex is painted in a color, and the color of Vertex i is c_i. Here, the color of each vertex is represented by an integer between 1 and N (inclusive). The same integer corresponds to the same color; different integers correspond to different colors.

For each k=1, 2, ..., N, solve the following problem:

Find the number of simple paths that visit a vertex painted in the color k one or more times.

Note: The simple paths from Vertex u to v and from v to u are not distinguished.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq c_i \leq N

1 \leq a_i,b_i \leq N

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
c_1 c_2 ... c_N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

Print the answers for k = 1, 2, ..., N in order, each in its own line.

Sample Input 1

3
1 2 1
1 2
2 3

Sample Output 1

5
4
0

Let P_{i,j} denote the simple path connecting Vertex i and j.

There are 5 simple paths that visit a vertex painted in the color 1 one or more times:

P_{1,1}\,,\,
P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,3}\,,\,
P_{3,3}

There are 4 simple paths that visit a vertex painted in the color 2 one or more times:

P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,2}\,,\,
P_{2,3}

There are no simple paths that visit a vertex painted in the color 3 one or more times.

Sample Input 2

1
1

Sample Output 2

1

Sample Input 3

2
1 2
1 2

Sample Output 3

2
2

Sample Input 4

5
1 2 3 4 5
1 2
2 3
3 4
3 5

Sample Output 4

5
8
10
5
5

Sample Input 5

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8

Sample Output 5

18
15
0
14
23
0
23
0

Solution in OCaml (4.10.0) (judge verdict: Accepted)

open Base
open Stdio

let read_line () =
    In_channel.(input_line_exn stdin)

let mapt2 ~f (a, b) = (f a, f b)

let array_replace arr i ~f = arr.(i) <- f (arr.(i))

let n = Int.of_string @@ read_line ()

let sums = Array.create ~len:(n+1) 0
let neg_res = Array.create ~len:(n+1) 0L
let g_hd = Array.create ~len:(n+1) 0
let g_nx = Array.create ~len:(2*n) 0
let g_to = Array.create ~len:(2*n) 0
let g_len = ref 0
let g_add u v =
    g_len := !g_len + 1;
    g_nx.(!g_len) <- g_hd.(u);
    g_to.(!g_len) <- v;
    g_hd.(u) <- !g_len
let g_adj_iter cur x0 f =
    let rec loop e acc = if e = 0 then acc else
        let ch = g_to.(e) in
        loop g_nx.(e) (f ch acc)
    in
    loop g_hd.(cur) x0

let colors = read_line () |> String.split ~on:' ' |> List.map ~f:Int.of_string
    |> List.cons 0 |> Array.of_list

let rec dfs ?(cur=1) ?(par=0) () =
    let memo = sums.(colors.(par)) in
    let tree_size = g_adj_iter cur 1 (fun ch acc ->
        acc + if ch = par then 0 else dfs ~cur:ch ~par:cur ())
    in

    array_replace sums colors.(cur) ((+) 1);
    let component_size = Int64.of_int @@ tree_size - (sums.(colors.(par)) - memo) in
    array_replace neg_res colors.(par) (fun x -> Int64.(x + component_size * (component_size + 1L) / 2L));
    sums.(colors.(par)) <- tree_size + memo;
    tree_size

let () =
    for i = 0 to n - 2 do
        let a, b = read_line () |> String.lsplit2_exn ~on:' ' |> mapt2 ~f:Int.of_string in
        g_add a b;
        g_add b a
    done

let () =
    ignore (dfs ());
    for i = 1 to n do
        let component_size = Int64.of_int @@ n - sums.(i) in
        let n64 = Int64.of_int n in
        let res = Int64.(n64 * (n64 + 1L) / 2L - component_size * (component_size + 1L) / 2L - neg_res.(i)) in
        printf "%Ld\n" res
    done