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 java.math.*
import kotlin.math.*

fun readLn()=readLine()!!
fun readInt()=readLn().toInt()
fun readInts()=readLn().split(" ").map{it.toInt()}
fun readLong()=readLn().toLong()
fun readLongs()=readLn().split(" ").map{it.toLong()}

val out=mutableListOf<String>()
fun printLine(s:String){out.add(s)}
fun output(){println(out.joinToString("\n"))}

fun ArrayList<Int>.lower_bound(x:Int):Int{
    var i=this.binarySearch(x)
    if(i>=0)return i
    return -i-1
}

fun main(){
    var n=readInt()
    var a=readInts().toIntArray()
    var E=Array(n){ArrayList<Int>()}
    for(i in 1 until n){
        var (u,v)=readInts()
        u--;v--
        E[u].add(v)
        E[v].add(u)
    }
    var ans=Array(n){0}
    var lis=ArrayList<Int>()
    fun DFS(u:Int,p:Int){
        var pop=false
        var i=lis.lower_bound(a[u])
        var las=0
        if(i==lis.size){
            pop=true
            lis.add(a[u])
        }else{
            las=lis[i]
            lis[i]=a[u]
        }
        ans[u]=lis.size
        for(v in E[u])if(v!=p)DFS(v,u)
        if(pop)lis.removeAt(lis.size-1)
        else lis[i]=las
    }
    DFS(0,-1)
    for(i in 0 until n)printLine("${ans[i]}")
    output()
}