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() {
  int N;
  cin >> N;
  vector<int> A(N);
  for(auto& i: A) cin >> i;
  vector<vector<int>> G(N);
  for(auto i=1; i<N; ++i) {
    int u, v;
    cin >> u >> v;
    --u; --v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }

  vector<int> d(N, 1);
  vector<int> dp = {A[0]};
  function<void(int, int)> dfs = [&](auto pre, auto cur) {
    auto i = -1, p = -1;
    if(dp.back() < A[cur]) {
      dp.emplace_back(A[cur]);
    } else {
      i = lower_bound(begin(dp), end(dp), A[cur]) - begin(dp);
      p = dp[i];
      dp[i] = A[cur];
    }
    d[cur] = dp.size();
    for(auto nex: G[cur]) if(pre != nex) dfs(cur, nex);
    if(i < 0) dp.pop_back();
    else      dp[i] = p;
  };

  dfs(0, 0);

  for(auto i: d) cout << i << endl;
}