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::{fastout, input};
struct Prob {
    a: Vec<usize>,
    g: Vec<Vec<usize>>,
    ans: Vec<usize>,
    dp: Vec<usize>,
}

impl Prob {
    fn dfs<'a>(&'a mut self, pos: usize, par: usize, len: usize) {
        let mut lb = 0;
        let mut ub = len;
        while ub - lb > 1 {
            let mid = (lb + ub) / 2;
            let ok = self.dp[mid] < self.a[pos];
            if ok {
                lb = mid;
            } else {
                ub = mid;
            }
        }
        let j = lb + 1;
        let prev_dp_j = self.dp[j];
        let prev_len = len;
        self.dp[j] = self.dp[j].min(self.a[pos]);
        let len = len.max(j + 1);
        self.ans[pos] = len - 1;
        for j in 0..self.g[pos].len() {
            let k = self.g[pos][j];
            if k == par {
                continue;
            }
            self.dfs(k, pos, len);
        }
        self.dp[j] = prev_dp_j;
    }
}

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

    let mut g: Vec<Vec<usize>> = vec![vec![]; n];
    for (u, v) in e {
        g[u - 1].push(v - 1);
        g[v - 1].push(u - 1);
    }

    let mut p = Prob {
        a,
        g,
        dp: vec![usize::max_value(); n + 1],
        ans: vec![usize::max_value(); n + 1],
    };
    p.dp[0] = 0;
    p.dfs(0, n, 1);
    for i in 0..n {
        println!("{}", p.ans[i]);
    }
}