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;
typedef long long ll;

const int nax = 2e5;
vector<int> node[nax];
int ans[nax];
int a[nax];

vector<int> cur;
void dfs(int p, int par) {
  int j = lower_bound(cur.begin(), cur.end(), a[p])-cur.begin();
  int added = (j == (int)cur.size()), cp;
  if (added)
    cur.push_back(a[p]);
  else {
    cp = cur[j];
    cur[j] = a[p];
  }

  ans[p] = cur.size();
  for (int i : node[p]) {
    if (i == par) continue;
    dfs(i, p);
  }
  if (added) cur.pop_back();
  else cur[j] = cp;
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].push_back(b);
    node[b].push_back(a);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++)
    cout << ans[i] << '\n';
}
