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;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(obj) (obj).begin(),(obj).end()
typedef pair<int,int> P;
const int mod=1e9+7;
vector<int> dp,ans,a,g[200005];
void dfs(int v,int p=-1){
  int i=lower_bound(all(dp),a[v])-dp.begin();
  int old=dp[i];
  dp[i]=a[v];
  ans[v]=i;
  if(p!=-1)ans[v]=max(ans[v],ans[p]);
  for(auto u:g[v]){
    if(u==p)continue;
    dfs(u,v);
  }
  dp[i]=old;
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;cin>>n;
  a=ans=vector<int>(n);
  rep(i,n)cin>>a[i];
  rep(i,n-1){
      int u,v;cin>>u>>v;
      --u;--v;
      g[u].push_back(v);
      g[v].push_back(u);
  }
  dp=vector<int>(n+1,mod);
  dp[0]=-1;
  dfs(0);
  rep(i,n)cout<<ans[i]<<endl;
  return 0;
  }
