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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::*;
use proconio::marker::Usize1;

#[fastout]
fn run() {
    input! {
        n: usize,
        a: [u32; n],
        e: [(Usize1, Usize1); n - 1],
    }
    let mut g = vec![vec![]; n];
    for (a, b) in e {
        g[a].push(b);
        g[b].push(a);
    }
    let mut ans = vec![0; n];
    enum Operation {
        DFS(usize, usize),
        ROLLBACK(usize, u32),
        POP,
    }
    let mut dp = vec![0];
    let mut stack = vec![Operation::DFS(0, 0)];// 頂点、親、命令
    while let Some(op) = stack.pop() {
        match op {
            Operation::DFS(v, p) => {
                let mut l = 0;
                let mut r = dp.len();
                while r - l > 1 {
                    let m = (l + r) / 2;
                    if dp[m] < a[v] {
                        l = m;
                    } else {
                        r = m;
                    }
                }
                if r >= dp.len() {
                    dp.push(a[v]);
                    stack.push(Operation::POP);
                } else {
                    let save = dp[r];
                    dp[r] = a[v];
                    stack.push(Operation::ROLLBACK(r, save));
                }
                ans[v] = dp.len() - 1;
                for &u in g[v].iter() {
                    if u != p {
                        stack.push(Operation::DFS(u, v));
                    }
                }
            },
            Operation::ROLLBACK(v, val) => {
                dp[v] = val;
            },
            Operation::POP => {
                dp.pop();
            }
        }
    }
    for a in ans {
        println!("{}", a);
    }
}

fn main() {
    run();
}