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 int LL;
typedef pair<int,int> P;
typedef pair<LL,LL> PL;

const int MOD=1000000007;
const int INF=1001001001;
const LL LINF=100100100100100;

LL n;
LL a[200010];
vector<LL> g[200010];
LL dp[200010];
LL ans[200010];

void dfs(LL v,LL p){
  LL ima=lower_bound(dp,dp+200010,a[v])-dp;
  LL old=dp[ima];
  dp[ima]=a[v];
  ans[v]=lower_bound(dp,dp+200010,LINF)-dp-1;
  for(LL w:g[v]){
    if(w==p)continue;
    dfs(w,v);
  }
  dp[ima]=old;
}
  
int main(){
  cin>>n;
  for(LL i=0;i<n;i++)cin>>a[i];
  for(LL i=0;i<n-1;i++){
    LL u,v;cin>>u>>v;u--;v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for(LL i=0;i<200010;i++)dp[i]=LINF;
  dp[0]=-LINF;
  dfs(0,-1);
  for(LL i=0;i<n;i++)cout<<ans[i]<<endl;
  return 0;
}