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 Kotlin (1.3.71) (judge verdict: Runtime Error)

import java.io.BufferedReader
import java.io.InputStreamReader
import java.util.*
import kotlin.collections.HashSet


fun main() {
    val bd = BufferedReader(InputStreamReader(System.`in`))
//    val sc = Scanner(System.`in`)
    val n = bd.readLine()!!.toInt()

    val colors = IntArray(n + 1)
    val ans = LongArray(n + 1)
    val colorCount = IntArray(n + 1)
    val childCount = IntArray(n + 1)
    val visited = BooleanArray(n + 1)
    val edge = Array(n+1){HashSet<Int>()}

    val bc = bd.readLine()!!.split(" ").map { it.toInt() }
    for(i in 1..n) {
        colors[i] = bc[i - 1]
        ans[colors[i]]++
    }

    for(i in 1..n) {
        val minus = (ans[i] * (ans[i] - 1L))/ 2L
        val plus = ans[i] * n.toLong()
        ans[i] = plus - minus
    }

    for(i in 1 until n) {
        val (f,t) = bd.readLine()!!.split(" ").map { it.toInt() }
        edge[f].add(t)
        edge[t].add(f)
    }

    var root = 1

    for(i in 1..n) {
        if(edge[i].size == 1) {
            root = i
            break
        }
    }

    fun dfs(v:Int, p:Int) : Int{
        if(visited[v]) return 0
        val color = colors[v]
        colorCount[color]++
        visited[v] = true

        val startColorCount = colorCount[color]
        val startChildCount = childCount[color]
        var curColorCount = startColorCount
        var curChildCount = startChildCount
        var totalChildCount = 0
        var realCount = 0
        val list = ArrayList<Long>()

        if(edge[v].isNotEmpty()) {
            for(child in edge[v]) {
                if(child == p) continue
                val childNum = dfs(child, v)
                totalChildCount += childNum

                val diffColor = colorCount[color] - curColorCount
                val diffChild = childCount[color] - curChildCount
                val cnt = childNum - diffChild - diffColor
                list.add(cnt.toLong())

                realCount += childNum - diffChild - diffColor
                curColorCount = colorCount[color]
                curChildCount = childCount[color]
            }
        }
        list.add((n - totalChildCount - startChildCount - startColorCount).toLong())
        val sum = list.sum()
        ans[color] += sum * (sum - 1L) / 2L

        for(d in list) {
            val x = d
            ans[color] -= x * (x - 1L) / 2L
        }
        childCount[color] += realCount
        return totalChildCount + 1
    }

    dfs(1, -1)
    for(i in 1..n) {
        println(ans[i])
    }
}