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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    vector<int> a(n);
    for (auto &x: a) cin >> x;
    vector<int> ans(n);
    int inf = (int) 1e9 + 20;
    vector<int> lis(n+1, inf); lis[0] = 0;
    vector<array<int, 2>> pr(n);
    vector<vector<int>> e(n);
    for (int i = 1; i < n; i++) {
        int u, v; cin >> u >> v; u--, v--;
        e[u].emplace_back(v);
        e[v].emplace_back(u);
    }
    int len = 0;
    auto f = [&](auto f, int curr, int prev) -> void {
        int x = a[curr];
        int i = lower_bound(lis.begin(), lis.end(), x) - lis.begin();
        pr[curr] = {i, lis[i]};
        lis[i] = x; 
        if (len < i) len = i;
        ans[curr] = len;
        for (auto &to: e[curr]) {
            if (prev == to) continue;
            f(f, to, curr);
        }
        lis[pr[curr][0]] = pr[curr][1];
        if (pr[curr][1] == inf) len--;
    };
    f(f, 0, -1);
    for (int i = 0; i < n; i++) cout << ans[i] << "\n";
    return 0;
}