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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
constexpr int N = 202020;
vector<int> g[N];
int a[N];
int v[N];
stack<int> st[N];
int updated[N];
int ans[N];
void dfs (int cur, int par) {
  int p = lower_bound(v, v + N, a[cur]) - v;
  st[p].push(v[p]);
  v[p] = a[cur];
  ans[cur] = lower_bound(v, v + N, 1 << 30) - v;
  for (int child : g[cur]) if (child != par) {
    dfs(child, cur);
  }
  v[p] = st[p].top();
  st[p].pop();
}
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  fill(v, v + N, 1 << 30);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}