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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int n;
vector<int> a;
vector<vector<int>> e;
vector<int> dp;
vector<int> ans;

void dfs(int pre, int crt, int size)
{
  auto ud = lower_bound(dp.begin(), dp.begin() + size + 1, a[crt]);
  int tmp = *ud;
  *ud = a[crt];
  if (tmp == INT_MAX)
    size++;
  ans[crt] = size;
  for (int i : e[crt])
    if (i != pre)
      dfs(crt, i, size);
  *ud = tmp;
}

int main()
{
  cin >> n;
  a.resize(n);
  rep(i, n)
    cin >> a[i];
  e.resize(n);
  rep(i, n - 1)
  {
    int u, v;
    cin >> u >> v;
    e[u-1].push_back(v - 1);
    e[v-1].push_back(u - 1);
  }
  ans.resize(n);
  dp.resize(n);
  fill(dp.begin(), dp.end(), INT_MAX);
  dfs(-1, 0, 0);
  rep(i, n)
    printf("%d\n", ans[i]);
  return 0;
}