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;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default)]
struct Struct;

// [l,h) の中でもっとも手前のfalseになる場所を探す. 全てのtrueはfalseより手前にある必要がある.
fn partition_point<F>(l0: usize, h0: usize, f: F) -> usize
where
    F: Fn(usize) -> bool,
{
    let mut l = l0;
    let mut h = h0;

    while h - l > 0 {
        let m = l + (h - l) / 2;
        if f(m) {
            l = m + 1;
        } else {
            h = m;
        }
    }
    return l;
}

const MAX_A: usize = 1000_000_001;

fn dfs(
    a: &Vec<usize>,
    v: &Vec<Vec<usize>>,
    i: usize,
    parent: usize,
    dp: &mut Vec<usize>,
    ans: &mut Vec<usize>,
) {
    let pos = partition_point(0, dp.len(), |pos| dp[pos] < a[i]);
    if pos == dp.len() {
        dp.push(MAX_A);
    }
    let last_val = dp[pos];
    dp[pos] = a[i];
    
    ans[i] = dp.len()-1;
    for &next_i in &v[i] {
        if next_i != parent {
            dfs(a, v, next_i, i, dp, ans);
        }
    }
    
    dp[pos] = last_val;
    if last_val == MAX_A {
        dp.pop();
    }
}

//#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        a: [usize;n],
        uv: [(Usize1,Usize1);n-1],
    }

    let mut v = vec![Vec::new(); n];
    for x in &uv {
        v[x.0].push(x.1);
        v[x.1].push(x.0);
    }

    let mut dp = Vec::new();
    dp.push(0);
    let mut ans = vec![0; n];

    dfs(&a, &v, 0, n, &mut dp, &mut ans);
    for i in 0..n {
        println!("{}", ans[i]);
    }
}