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::cmp::min;

const INF: u64 = 1 << 50;

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let n: usize = sc.read();
    let a: Vec<u64> = sc.vec(n);
    let mut graph = vec![vec![]; n];
    for _ in 1..n {
        let u = sc.read::<usize>() - 1;
        let v = sc.read::<usize>() - 1;
        graph[u].push(v);
        graph[v].push(u);
    }

    let mut memo = vec![INF; n + 2];
    memo[0] = 0;
    memo[1] = a[0];
    let mut ans = vec![0; n];
    ans[0] = 1;
    dfs(0, 0, &graph, &a, &mut memo, &mut ans);
    for ans in ans.into_iter() {
        println!("{}", ans);
    }
}

fn dfs(
    v: usize,
    p: usize,
    graph: &Vec<Vec<usize>>,
    a: &Vec<u64>,
    memo: &mut Vec<u64>,
    ans: &mut Vec<usize>,
) {
    for &next in graph[v].iter() {
        if p == next {
            continue;
        }
        let a_next = a[next];
        let prev = memo
            .binary_search_by_key(&(a_next * 2), |&m| m * 2 + 1)
            .err()
            .unwrap()
            - 1;
        assert!(memo[prev] < a_next);
        assert!(memo[prev + 1] >= a_next);

        let old = memo[prev + 1];
        memo[prev + 1] = min(memo[prev + 1], a_next);

        ans[next] = memo
            .binary_search_by_key(&(INF * 2), |&m| m * 2 + 1)
            .err()
            .unwrap()
            - 1;

        dfs(next, v, graph, a, memo, ans);
        memo[prev + 1] = old;
    }
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> Self {
        Self(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.to_string().as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}