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;
using ll = long long;

const int INF = 2e9;

void dfs(int x, int bef, int dp[], const int n, const vector<int> &A, const vector<vector<int>> &T, int ans[]) {
  int *p = lower_bound(dp ,dp+n+1, A[x]);
  int tmp = *p;
  *p = min(tmp, A[x]);
  ans[x] = lower_bound(dp ,dp+n+1, INF) - dp;

  for (auto t: T[x]) {
    if (t == bef) continue;

    dfs(t, x, dp, n, A, T, ans);
  }
  *p = tmp;

  return;
}

int main() {
  int n;
  cin >> n;

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

  int dp[n+1];
  fill(dp, dp+n+1, INF);
  dp[0] = A[0];

  int ans[n];

  dfs(0, -1, dp, n ,A, T, ans);

  for (int i = 0; i < n; i++) {
    cout << ans[i] << endl;
  }

  return 0;
}