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 ans[223456],dp[223456],v[223456],n,INF=1e9+1,nowans;
vector<int>edge[223456];

void dfs(int parent,int idx){
  int l=0,r=n+1;
  while(r-l>1){
    int mid=(l+r)>>1;
    (dp[mid]<v[idx]?l:r)=mid;
  }
  int preidx=r,prevalue=dp[r];
  dp[r]=v[idx];
  if(nowans<r)nowans=r;
  ans[idx]=nowans;
  for(int q:edge[idx])if(q!=parent)dfs(idx,q);
  dp[preidx]=prevalue;
  while(dp[nowans]==INF)nowans--;
}

int main(){
  for(int i=1;i<223456;i++)dp[i]=INF;
  cin>>n;
  for(int i=0;i<n;i++)cin>>v[i];
  for(int i=1;i<n;i++){
    int a,b;cin>>a>>b;a--;b--;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  dfs(-1,0);
  for(int i=0;i<n;i++)cout<<ans[i]<<endl;
}
