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: Wrong Answer)

#ifndef DEBUG
#pragma GCC optimize("O3,unroll-loops")
#pragma loop_opt(on)
#endif
#include <bits/stdc++.h>
using namespace std;

#ifdef DEBUG
#define FAST ;
#else
#define FAST ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
#define cerr
#endif
#define _ <<' '<<
#define Fr(i, s, e) for(auto i = s; i < e; ++i)
#define All(v) v.begin(),v.end()
#define Ft first
#define Sd second
using ll = long long;
using pii = pair<int, int>;
constexpr ll MOD = 998'244'353;

signed main() {
    FAST;

    int n;
    cin >> n;
    vector<int> a(n);
    for(int &x: a) cin >> x;
    vector<vector<int>> edge(n);
    Fr(i, 1, n) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        edge[u].emplace_back(v);
        edge[v].emplace_back(u);
    }
    vector<int> ans(n);
    function<void(int,int)> dfs;
    dfs = [&](int now, int par) {
        static vector<int> LCS{ a[now] };
        //cerr _ par _ now _ ": "; for(int x: LCS)  cerr _ x; cerr _ endl;
        ans[now] = LCS.size();
        for(int x: edge[now]) if (x != par) {
            size_t idx = distance(LCS.begin(), lower_bound(All(LCS),a[x]+1));
            int back = -1;
            if (idx == LCS.size()) {
                back = -1;
                LCS.emplace_back(a[x]);
            } else {
                back = LCS[idx];
                LCS[idx] = a[x];
            }
            dfs(x, now);
            if (back == -1) LCS.pop_back();
            else LCS[idx] = back;
        }
    };
    dfs(0,-1);
    Fr(i, 0, n) cout << ans[i]
#ifdef DEBUG
        << " \n"[i+1==n];
#else
        << endl;
#endif

}