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>
#define rep(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
using ll = long long;
const int INF = (1<<30)-1;
const long long LINF = (1LL<<62)-1;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

vector<vector<int>> to;
vector<int> a, res;
void dfs(int now, int prev, vector<int> &lis) {
  auto it = upper_bound(lis.begin(), lis.end(), a[now]);
  int past = *it;
  *it = a[now];
  res[now] = max(res[prev], int(it-lis.begin()+1));
  for (auto next : to[now]) {
    if (next == prev)
      continue;
    dfs(next, now, lis);
  }
  *it = past;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  a.resize(n);
  rep(i, n)
    cin >> a[i];
  to.resize(n);
  rep(i, n-1) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    to[u].push_back(v);
    to[v].push_back(u);
  }
  res.resize(n);
  vector<int> lis(n, INF);
  dfs(0, 0, lis);
  rep(i, n)
    cout << res[i] << '\n';
  return 0;
}