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;
typedef pair<ll,ll> P;
const ll INF=1e18;
#define rep(i, n) for (ll i = 0; i < (int)(n); i++)

vector<ll> a(200005);
vector<ll> to[200005];
vector<ll> dp(200005,INF);
vector<ll> ans(200005,0);


int dfs(int t, int p){
  ll id=lower_bound(dp.begin(),dp.end(),a[t])-dp.begin();
  ll prev=dp[id];
  dp[id]=a[t];
  if (p!=-1) ans[t]=max(id,ans[p]);
  else ans[t]=id;
//  cout<<t<<' '<<id<<' '<<a[t]<<endl;
  for(int u:to[t]){
    if (u==p)continue;
    dfs(u,t);
  }
  dp[id]=prev;
  return 0;

}
int main(){
  ll n;
  cin>>n;


  rep(i,n){
    cin>>a[i];
  }
  rep(i,n-1){
    ll u1,v1;
    cin>>u1>>v1;
    u1--;v1--;
    to[u1].push_back(v1);
    to[v1].push_back(u1);
  }
  dp[0]=-INF;
  dfs(0,-1);
  rep(i,n)cout<<ans[i]<<endl;

}
