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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn dfs(
    cur: usize,
    parent: usize,
    a: &Vec<i64>,
    graph: &Vec<Vec<usize>>,
    dp: &mut Vec<i64>,
    answers: &mut Vec<usize>,
) {
    let pointer = dp.lower_bound(&a[cur]);
    let prev = dp[pointer];
    dp[pointer] = a[cur];
    answers[cur] = dp.lower_bound(&std::i64::MAX);
    for &to in graph[cur].iter() {
        if to == parent {
            continue;
        }
        dfs(to, cur, a, graph, dp, answers);
    }
    dp[pointer] = prev;
}

fn main() {
    let n = read::<usize>();
    let a = read_vec::<i64>();
    let mut graph = vec![vec![]; n];
    for i in 0..n - 1 {
        let v = read_vec::<usize>();
        let (a, b) = (v[0] - 1, v[1] - 1);
        graph[a].push(b);
        graph[b].push(a);
    }
    let mut dp = vec![std::i64::MAX; n + 1];
    let mut answers = vec![std::usize::MAX; n];
    dfs(0, 0, &a, &graph, &mut dp, &mut answers);
    for ans in answers {
        println!("{}", ans);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
pub trait BinarySearch<T> {
    fn lower_bound(&self, x: &T) -> usize;
}

impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less => {
                    low = mid + 1;
                }
                Ordering::Equal | Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
}