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;
#define ll long long
#define endl '\n'
#define sz(x) (int)x.size() 
void dbg_out() { cerr << "\b\b]\n"; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T){ cerr << H << ", "; dbg_out(T...);}
#define watch(...) cerr << "[" << #__VA_ARGS__ << "]: [", dbg_out(__VA_ARGS__)


/****************************** CODE IS HERE ****************************************/
const int N = 2e5 + 7;
vector <int> tree[N];
int A[N];
vector <vector <int> > seq;
bitset<N> used;
vector <int> ans(N);

int bi_search(int x){
    int lo = 0, hi = sz(seq)-1;
    if(hi < 0) return 0;
    while(lo < hi){
      int mid = (lo + hi + 1) >> 1;
      if(seq[mid].back() <= x)
        lo = mid;
      else hi = mid-1;
    }
    return (seq[lo].back() > x ? lo: lo+1);
}


void dfs(int u = 1, int p = 0){
  int idx = bi_search(A[u]);
  if(idx >= sz(seq)){
    seq.push_back({A[u]});
  }
  else seq[idx].push_back(A[u]);
  ans[u] = idx + 1;
  // watch(u, idx+1);
  // watch(u);
  // for (int i = 0; i < sz(seq); ++i)
    // for (int j = 0; j < sz(seq[i]); ++j)
      // cout << seq[i][j] << " \n"[j==sz(seq[i])-1];
  for (int v: tree[u]){
    if(v == p) continue;
    dfs(v, u);
  }
  seq[idx].pop_back();
  if(sz(seq[idx]) == 0)
    seq.pop_back();
}

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

    int n; cin >> n;
    for (int i = 1; i <= n; ++i)
      cin >> A[i];
    for (int i = 1, x, y; i < n; ++i){
      cin >> x >> y;
      tree[x].push_back(y);
      tree[y].push_back(x);
    }

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

  return 0;
}