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>
typedef long long ll;
using namespace std;
vector<vector<int>> adj;
vector<int> vs;
vector<int> lis;
vector<int> ans;
int n;

int get_index(int u){
  int lo = 0, hi = n - 1;
  while(lo < hi){
    int mid = (lo + hi) / 2;
    if(lis[mid] >= vs[u])
      hi = mid;
    else
      lo = mid + 1;
  }
  return lo;
}

void dfs(int u, int p = -1){
//cout << u << endl;
  int lis_in = get_index(u);
  int old_v = lis[lis_in];
  lis[lis_in] = vs[u];
  ans[u] = lis_in + 1;
  if(p != -1)
    ans[u] = max(ans[u], ans[p]);
  for(auto el : adj[u]){
    if(el == p)continue;
    dfs(el, u);
  }
  lis[lis_in] = old_v;
}

int main()
{
  cin >> n;
  adj.resize(n);
  vs.resize(n);
  ans.resize(n);
  lis.resize(n + 1, 1e9 + 1);
  for(int i = 0;i < n;i++)
    cin >> vs[i];
  for(int i = 0;i < n - 1;i++){
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }

  dfs(0);
  for(auto el : ans)
    cout << el << endl;
  return 0;
}