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;

vector<int> T;
vector<int> seen;
vector<vector<int>> G;
stack<pair<int,int>> S;
vector<int> p(200000,1000000001);
void DFS(int a){
  int b=lower_bound(p.begin(),p.end(),T.at(a))-p.begin(),c=p.at(b);
  p.at(b)=T.at(a);
  seen.at(a)=lower_bound(p.begin(),p.end(),1000000001)-p.begin();
  S.push(make_pair(b,c));
  for(int x:G.at(a)){
    if(seen.at(x)!=0){
      continue;
    }
    else{
      DFS(x);
    }
  }
  int B=S.top().first,C=S.top().second;
  S.pop();
  p.at(B)=C;
}
  

int main() {
  int N;
  cin>>N;
  T.resize(N);
  seen.resize(N);
  G.resize(N);
  for(int i=0;i<N;i++){
    cin>>T.at(i);
  }
  for(int i=0;i<N-1;i++){
    int a,b;
    cin>>a>>b;
    a--;
    b--;
    G.at(a).push_back(b);
    G.at(b).push_back(a);
  }
  DFS(0);
  for(int i=0;i<N;i++){
    cout<<seen.at(i)<<endl;
  }
}