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 int64_t ll;
int a[200001],dp[200001],ans[200001];
vector<int> Graph[200001];

void dfs(int now,int parent,int mx){
  int ng=0,ok=mx,mid;
  while(ok-ng>1){
    mid = ng + (ok - ng)/2;
    if(dp[mid]<a[now]) ng = mid;
    else ok = mid;
  }
  int save = dp[ok];
  dp[ok] = a[now];
  ans[now] = (ok==mx?mx++:mx-1);
  for(int child:Graph[now]) if(child != parent) dfs(child,now,mx);
  dp[ok] = save;
}

int main() {
  int n,u,v;
  cin >> n;
  for(int i=1;i<=n;i++) cin >> a[i];
  for(int i=1;i<=n-1;i++){
    cin >> u >> v;
    Graph[u].push_back(v);
    Graph[v].push_back(u);
  }
  for(int i=1;i<=n;i++) dp[i]=2e9;
  dfs(1,-1,1);
  for(int i=1;i<=n;i++) cout << ans[i] << endl;
}
