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 <iostream>
#include <vector>
#include <iterator>
using namespace std;
constexpr int INF = 1e9 + 10;

int par[200010], a[200010], ans[200010];
vector<int> G[200010];
void dfs(int p, vector<int> &dp)
{
  for (auto u : G[p])
  {
    if (par[p] == u)
      continue;
    par[u] = p;
    auto itr = lower_bound(dp.begin(), dp.end(), a[u]);
    int before = *itr;
    *itr = a[u];
    auto INF_L = lower_bound(dp.begin(), dp.end(), INF);
    ans[u] = distance(dp.begin(), INF_L);
    dfs(u, dp);
    *itr = before;
  }
}

int main()
{
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 0; i < n - 1; i++)
  {
    int u, v;
    cin >> u >> v;
    u--, v--;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }

  vector<int> dp(n, INF);
  dp[0] = a[0];
  ans[0] = 1;
  dfs(0, dp);
  for (int i = 0; i < n; i++)
    cout << ans[i] << "\n";
  return 0;
}
