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::input;
use proconio::marker::Usize1;
use superslice::*;

fn dfs(
    ls: &mut Vec<usize>,
    lis: &mut Vec<u64>,
    es: &Vec<Vec<usize>>,
    ais: &Vec<u64>,
    u: usize,
    p: usize,
) {
    let a = ais[u];
    let b = if *(lis.last().unwrap_or(&0)) < a {
        lis.push(a);
        None
    } else {
        let i = lis[..].lower_bound(&a);
        let x = lis[i];
        lis[i] = a;
        Some((i, x))
    };
    ls[u] = lis.len();
    for &v in es[u].iter() {
        if v == p {
            continue;
        }
        dfs(ls, lis, es, ais, v, u);
    }
    match b {
        None => {
            lis.pop();
        }
        Some((i, x)) => lis[i] = x,
    }
}

fn main() {
    input! {
        n: usize,
        ais: [u64; n],
        uvs: [(Usize1, Usize1); n - 1],
    };

    let mut es = vec![vec![]; n];
    for (u, v) in uvs {
        es[u].push(v);
        es[v].push(u);
    }

    let mut ls = vec![0_usize; n];
    dfs(&mut ls, &mut vec![], &es, &ais, 0, 0);

    for l in ls {
        println!("{}", l);
    }
}