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)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let a: Vec<i32> = s.by_ref().take(n).map(|a| a.parse().unwrap()).collect();
	let mut edges = vec![vec![]; n];
	for _ in 0..n - 1 {
		let a = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let b = s.next().unwrap().parse::<usize>().unwrap() - 1;
		edges[a].push(b);
		edges[b].push(a);
	}
	let mut ans = vec![0; n];
	let mut dp = vec![std::i32::MAX; n];
	let graph = Graph { edges, a };
	graph.dfs(&mut ans, &mut dp, 0, std::usize::MAX, 0);
	use std::io::Write;
	let out = std::io::stdout();
	let mut out = out.lock();
	for ans in ans {
		writeln!(out, "{}", ans).unwrap();
	}
}

struct Graph {
	edges: Vec<Vec<usize>>,
	a: Vec<i32>,
}

impl Graph {
	pub fn dfs(&self, ans: &mut [i32], dp: &mut [i32], v: usize, last: usize, depth: usize) {
		let i = dp[..depth].lower_bound(&self.a[v]);
		ans[v] = i as i32 + 1;
		let l = dp[i];
		dp[i] = self.a[v];
		for &to in self.edges[v].iter() {
			if to != last {
				self.dfs(ans, dp, to, v, depth + 1);
			}
		}
		dp[i] = l;
	}
}

pub trait BinarySearch<'a, T: 'a> {
	fn lower_bound(&self, x: &T) -> usize;
}

impl<'a, T: 'a + Ord> BinarySearch<'a, T> for [T] {
	fn lower_bound(&self, x: &T) -> usize {
		let mut first = 0;
		let mut len = self.len() - first;
		while len > 0 {
			let half = len >> 1;
			let mid = first + half;
			use std::cmp::Ordering;
			match self[mid].cmp(x) {
				Ordering::Less => {
					first = mid + 1;
					len = len - half - 1;
				}
				Ordering::Equal | Ordering::Greater => {
					len = half;
				}
			}
		}
		first
	}
}