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;

int n;
int a[200000];
vector<int> g[200000];
int ans[200000], dp[200000];
stack<pair<int,int>> s;

void dfs(int cur, int p) {
  auto it = lower_bound(dp, dp + n, a[cur]);
  int idx = it - dp;
  s.push(make_pair(idx, dp[idx]));
  dp[idx] = min(dp[idx], a[cur]);
  ans[cur] = lower_bound(dp, dp + n, 1e9+1) - dp;
  for (int to: g[cur]) {
    if (to == p) {
      continue;
    }
    dfs(to, cur);
  }
  auto top = s.top();
  s.pop();
  dp[top.first] = top.second;
}
int main() {
  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].push_back(v);
    g[v].push_back(u);
  }
  for (int i=0; i<n; i++) ans[i] = 0;
  for (int i=0; i<n; i++) dp[i] = 1e9 + 1;
  dfs(0, -1);
  for (int i=0; i<n; i++) {
    cout<<ans[i]<<endl;
  }
}