Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices, whose i-th edge connects Vertex u_i and Vertex v_i.
Vertex i has an integer a_i written on it.
For every integer k from 1 through N, solve the following problem:

We will make a sequence by lining up the integers written on the vertices along the shortest path from Vertex 1 to Vertex k, in the order they appear. Find the length of the longest increasing subsequence of this sequence.

Here, the longest increasing subsequence of a sequence A of length L is the subsequence A_{i_1} , A_{i_2} , ... , A_{i_M} with the greatest possible value of M such that 1 \leq i_1 < i_2 < ... < i_M \leq L and A_{i_1} < A_{i_2} < ... < A_{i_M}.

Constraints

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

1 \leq a_i \leq 10^9

1 \leq u_i , v_i \leq N

u_i \neq v_i

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
u_1 v_1
u_2 v_2
:
u_{N-1} v_{N-1}

Output

Print N lines. The k-th line, print the length of the longest increasing subsequence of the sequence obtained from the shortest path from Vertex 1 to Vertex k.

Sample Input 1

10
1 2 5 3 4 6 7 3 2 4
1 2
2 3
3 4
4 5
3 6
6 7
1 8
8 9
9 10

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in Kotlin (1.3.71) (judge verdict: Runtime Error)

import java.util.*
import kotlin.collections.ArrayList

val sc = Scanner(System.`in`)
fun nextIntArray() = sc.nextLine().split(" ").map(String::toInt).toTypedArray()

const val INF = 10000000

var N = 0
var a = Array(0){1}
var map = Array(0){ArrayList<Int>()}
// 最大部分増加列のダンプ
var dp = Array(0){0}
var cnt = 0
var ans = Array(0){0}

fun main(args: Array<String>) {

    N = sc.nextLine().toInt()
    a = nextIntArray()
    map = Array(N){ArrayList<Int>()}
    dp = Array(N){INF}
    ans = Array(N){0}

    for(i in 0..N-2){
        val u = sc.nextInt()-1
        val v = sc.nextInt()-1
        map[u].add(v)
        map[v].add(u)
    }

    calc(0, -1)

    for(t in ans){
        println(t)
    }
}

fun calc(i: Int, p: Int){
    // 自身のdpに自身の番号を適用する
    val idx = bisearch(dp, cnt, a[i])
    val pre = dp[idx]
    if(pre == INF) cnt++
    dp[idx] = a[i]
    ans[i] = cnt
    // 次の頂点へ再帰呼び出し
    for(j in map[i]){
        if(j == p) continue
        calc(j, i)
        dp[idx] = pre
        if(pre == INF)cnt--
    }
}

/**
 * 単調増加列arrの部分列[0]～[len-1]に対し、値ｖが挿入できる位置のインデックスを返す
 * ・二分探索
 * ・arr[len-1]<v　の場合は、lenが返る
 */
fun bisearch(arr: Array<Int>, len: Int, v: Int): Int{

    var left = 0
    var right = len
    var mid = 0
    while(left < right) {
        mid = (left + right) / 2
        if(arr[mid] < v){
            left = mid + 1
        }else if(arr[mid] > v){
            right = mid - 1
        }else {
            return mid
        }
    }

    // ここに来るパターンは２つ
    // １）最後のループで leftとrightの差が1しかなく（left=mid）、right=mid-1の結果、right<leftになってしまった
    if(right < left){
        return left
    }
    // 2)leftかrightの再計算により、left=rightとなった時
    else{
        return if(arr[left] < v) left + 1 else left
    }

}