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)

#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n; cin >> n;
    vector<int> a(n);
    for (auto& x: a) cin >> x;
    vector<vector<int>> g(n);
    for (int _ = 1; _ < n; _++) {
        int x,y;
        cin >> x >> y;
        --x;--y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    vector<int> dp(n+1, 2e9), len(n);
    dp[0] = 0;
    function<void(int,int)> dfs = [&](int u, int p){
        int k = lower_bound(dp.begin(), dp.end(), a[u]) - dp.begin();
        len[u] = k;
        int old = dp[k];
        dp[k] = a[u];
        for (int v: g[u])if(v!=p){
            dfs(v, u);
        }
        dp[k] = old;
    };
    dfs(0,-1);
    for (auto& x: len) {
        cout << x << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}