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;

typedef long long int ll;

const ll inf = 1e9;
const ll MAXN = 2e5 + 5;

vector < vector < ll > > adj(MAXN, vector < ll > ());
vector < ll > d(MAXN, inf), a(MAXN);
vector < ll > ans(MAXN, 1);

void dfs(ll u, ll p){
  for(auto to : adj[u]){
    if(to != p){
      ll tmp = -1;
      int j = upper_bound(d.begin(), d.end(), a[to]) - d.begin();
      if(a[to] > d[j - 1] && a[to] < d[j]){
        tmp = d[j];
        d[j] = a[to];
      }
      ll it = lower_bound(d.begin(), d.end(), inf) - d.begin();
      ans[to] = it - 1;
      dfs(to, u);
      if(tmp != -1) d[j] = tmp;
    }
  }
}

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  ll n;
  cin >> n;
  for(int i = 1; i <= n; i++) cin >> a[i];
  for(int i = 0; i < n - 1; i++){
    ll u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }

  d[0] = -inf;
  d[1] = a[1];
  dfs(1, -1);

  for(int i = 1; i <= n; i++) cout << ans[i] << endl;
}