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)

const val INF = 1_500_000_000
var n: Int = 0
var a: IntArray = intArrayOf()
var dp: IntArray = intArrayOf()
var ans: IntArray = intArrayOf()
var t: Array<MutableList<Int>> = arrayOf()

fun main(args: Array<String>) {
  n = readLine()!!.toInt()
  a = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
  t = Array<MutableList<Int>>(n) { mutableListOf<Int>() }
  repeat(n-1) {
    val (u, v) = readLine()!!.split(" ").map { it.toInt()-1 }
    t[u].add(v)
    t[v].add(u)
  }
  dp = IntArray(n+1) { INF }
  dp[0] = -1
  ans = IntArray(n)
  dfs(0, -1)
  ans.forEach { println(it) }
}
fun dfs(v: Int, p: Int) {
  val index = lowerBound(a[v])+1
  val prev = dp[index]
  dp[index] = a[v]
  ans[v] = lowerBound(INF-1)
  for (u in t[v]) {
    if (u == p) continue
    dfs(u, v)
  }
  dp[index] = prev
}
fun lowerBound(v: Int): Int {
  var l = 0
  var r = n+1
  while (r-l > 1) {
    val m = (l+r)/2
    if (dp[m] < v) l = m else r = m
  }
  return l
}