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::marker::Usize1;
use superslice::Ext;
static INF: i32 = std::i32::MAX;
fn main() {
    proconio::input! {
        n: usize,
        a: [i32;n],
        uv: [(Usize1,Usize1);n-1],
    }
    let mut g = vec![vec![];n];
    for (u,v) in uv {
        g[u].push(v);
        g[v].push(u);
    }
    let mut dp = vec![INF;n];
    let mut r = vec![0;n];
    f(0, &mut dp, &mut r, &a, &g);
    for r in r {
        println!("{}",r);
    }
}
fn f(v: usize, dp: &mut Vec<i32>, r: &mut Vec<usize>, a: &Vec<i32>, g: &Vec<Vec<usize>>) {
    let x = a[v];
    let k = dp.lower_bound(&x);
    let b = dp[k];
    dp[k]=x;
    r[v]=dp.lower_bound(&INF);
    for &u in &g[v] { if r[u]==0 {
        f(u, dp, r, a, g);
    }}
    dp[k]=b;
}