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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

package template.f;

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);

        int n  = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        List<Integer>[] adj = new List[n];
        for(int i=0; i<n; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        for(int i=0; i<n-1; i++) {
            int u = sc.nextInt()-1;
            int v = sc.nextInt()-1;
            adj[u].add(v);
            adj[v].add(u);
        }
        int[] lis = new int[n+1];
        int[] res = new int[n];
        Arrays.fill(lis, Integer.MAX_VALUE);
        lis[0] = 0;
        res[0] = 1;
        dfs(lis, res, a, adj,-1, 0);
        for(int i=0; i<n; i++) {
            System.out.println(res[i]);
        }
    }

    static void dfs(int[] lis, int[] res, int[] a, List<Integer>[] adj, int parent, int pos) {
        int pi = 0;
        int pv = 0;
        if(pos != 0) {
            pi = Arrays.binarySearch(lis, a[pos]);
            if(pi < 0) {
                pi = -pi - 1;
                pv = lis[pi];
                lis[pi] = a[pos];
            }
            int li = Arrays.binarySearch(lis, Integer.MAX_VALUE-1);
            li = -li - 2;
            res[pos] = li;
        } else {
            lis[1] = a[0];
        }
        for(int next : adj[pos]) {
            if(next == parent) continue;
            dfs(lis, res, a, adj, pos, next);
        }
        lis[pi] = pv;
    }

}