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: Wrong Answer)

macro_rules! input {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn dfs(start: usize, parent: usize, values: &Vec<usize>, edges: &Vec<Vec<usize>>, current_state: &mut Vec<usize>, lis: &mut Vec<usize>) {
    let longest_found = match current_state.binary_search(&values[start]) {
        Ok(a) => a,
        Err(a) => a,
    };
    let prev_value = current_state[longest_found];
    current_state[longest_found] = values[start];
    lis[start] = longest_found + 1;
    for &node in &edges[start] {
        if node != parent {
            dfs(node, start, values, edges, current_state, lis);
        }
    }
    current_state[longest_found] = prev_value;
}

trait VectorPrint {
    fn spaced_horizontal(&self) -> String;
    fn spaced_vertical(&self) -> String;
}

impl<T> VectorPrint for Vec<T> where T: std::fmt::Display {
    fn spaced_horizontal(&self) -> String {
        let mut s: String = self.iter().map(|x| format!("{} ", x)).collect();
        s.pop();
        s
    }
    fn spaced_vertical(&self) -> String {
        let mut s: String = self.iter().map(|x| format!("{}\n", x)).collect();
        s.pop();
        s
    }
}

fn main() {
    let stdin = std::io::stdin();
    let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
    let mut next = move || -> String{
        bytes
            .by_ref()
            .map(|r|r.unwrap() as char)
            .skip_while(|c|c.is_whitespace())
            .take_while(|c|!c.is_whitespace())
            .collect()
    };
    input! {next,
        n: usize,
        a: [usize; n],
        edge_read: [(usize1, usize1); n-1],
    }
    let mut edges = vec![Vec::new(); n];
    for (u,v) in edge_read {
        edges[u].push(v);
        edges[v].push(u);
    }
    let mut current_state = vec![std::usize::MAX; n];
    let mut lis = vec![0; n];
    dfs(0, 0, &a, &edges, &mut current_state, &mut lis);
    println!("{}", lis.spaced_vertical());
}