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: Time Limit Exceeded)

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>()}
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+1){ArrayList<Int>()}
    dp = Array(N){INF}
    ans = Array(N){0}

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

    val mae = Array(N+1){-1}
    val back_idx = Array(N+1){-1}
    val back_val = Array(N+1){-1}

    val stack = Stack<Int>()
    stack.push(1)
    while(stack.isNotEmpty()){

        var i = stack.pop()

        if(i < 0){
            i = -i
            dp[back_idx[i]] = back_val[i]
            if(back_val[i] == INF) cnt--
            continue
        }

        val s = dp.binarySearch(a[i-1], 0, cnt)
        val idx = if(s < 0) -s-1 else s

        val pre = dp[idx]
        if(pre == INF) cnt++
        dp[idx] = a[i-1]
        ans[i-1] = cnt

        stack.push(-i)
        back_idx[i] = idx
        back_val[i] = pre
        
        for(j in map[i]){
            if(mae[i] != j) {
                stack.push(j)
                mae[j] = i
            }
        }
    }
    
    for(t in ans){
        println(t)
    }
}