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 std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: ty),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

const M: usize = std::usize::MAX / 10;
fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize, a: [usize; n]);
    let mut g = vec![vec![]; n];
    for _ in 0..(n - 1) {
        parse!(it, a: usize, b: usize);
        g[a - 1].push(b - 1);
        g[b - 1].push(a - 1);
    }
    let mut ret = vec![0; n];
    let mut dp = vec![M; n];
    let mut seen = vec![false; n];
    seen[0] = true;
    dfs(&g, &mut ret, &mut dp, 0, &a, &mut seen);
    for r in ret {
        println!("{}", r);
    }
}
fn dfs(
    g: &Vec<Vec<usize>>,
    ret: &mut Vec<usize>,
    dp: &mut Vec<usize>,
    u: usize,
    a: &Vec<usize>,
    seen: &mut Vec<bool>,
) {
    let i = match dp.binary_search(&a[u]) {
        Ok(i) => i,
        Err(i) => i,
    };
    let previ = dp[i];
    dp[i] = a[u];
    ret[u] = dp.binary_search(&(M - 1)).unwrap_err();
    for &v in g[u].iter() {
        if !seen[v] {
            seen[v] = true;
            dfs(g, ret, dp, v, a, seen);
        }
    }
    dp[i] = previ
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}