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.

Example

Input

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


Output

1
2
3
3
4
4
5
2
2
3
#include <bits/stdc++.h>
using namespace std;

void dfs(const vector<vector<int>> &to, const vector<int> &a, vector<int> &lis, vector<int> &ans, int v, int p) {
  int index = lower_bound(lis.begin(), lis.end(), a.at(v)) - lis.begin();
  int old = lis.at(index);
  lis.at(index) = a.at(v);
  ans.at(v) = lower_bound(lis.begin(), lis.end(), 1e9) - lis.begin();
  for (int c : to.at(v)) {
    if (c == p) continue;
    dfs(to, a, lis, ans, c, v);
  }
  lis.at(index) = old;
}

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a.at(i);
  }
  vector<vector<int>> to(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--; v--;
    to.at(u).push_back(v);
    to.at(v).push_back(u);
  }
  vector<int> lis(n, 1e9), ans(n);
  dfs(to, a, lis, ans, 0, -1);
  for (int i = 0; i < n; i++) {
    cout << ans.at(i) << endl;
  }
}
