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)]
use proconio::{input, marker::*};

#[allow(unused_imports)]
use std::io::Write;
#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

const INF: usize = 1_000_000_000_000_000;

fn lower_bound<F>(l: isize, r: isize, criterion: F) -> isize
where
    F: Fn(isize) -> bool,
{
    let mut l = l;
    let mut r = r;
    while r - l > 1 {
        let mid = l + (r - l) / 2;
        if criterion(mid) {
            r = mid;
        } else {
            l = mid;
        }
    }
    r
}

fn dfs(
    node: usize,
    parent: usize,
    nodes: &Vec<Vec<usize>>,
    dp: &mut Vec<usize>,
    res: &mut Vec<usize>,
    a: &Vec<usize>,
    mut val: usize,
) {
    let crit = |i: isize| {
        let i = i as usize;
        a[node - 1] <= dp[i]
    };
    let i = lower_bound(-1, a.len() as isize, crit) as usize;
    if dp[i] == INF {
        val += 1;
    }
    let before = dp[i];
    dp[i] = a[node - 1];
    for &next in &nodes[node] {
        if next == parent {
            continue;
        }
        dfs(next, node, nodes, dp, res, a, val);
    }
    dp[i] = before;
    res[node - 1] = val;
}

fn main() {
    input! {
        n: usize,
        a: [usize; n],
        uv: [(usize,usize); n-1],
    }
    debug!(n, a, uv);

    let mut nodes = vec![Vec::new(); n + 1];
    for (u, v) in uv {
        nodes[u].push(v);
        nodes[v].push(u);
    }

    let mut res = vec![0; n];
    dfs(1, 0, &nodes, &mut vec![INF; n], &mut res, &a, 0);
    for i in 0..n {
        println!("{}", res[i]);
    }
}