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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int INF = 1e9;

void dfs(const vector<vector<int>>& G, const vector<int>& A, vector<int>& L, vector<int>& R, int v, int p){
  int lb = lower_bound(L.begin(), L.end(), A[v]) - L.begin();
  int old = L[lb];
  L[lb] = A[v];
  
  R[v] = lower_bound(L.begin(), L.end(), INF) - L.begin();
  for(auto v_ : G[v]){
    if(v_ == p) continue;
    dfs(G,A,L,R,v_,v);
  }
  L[lb] = old;
}

int main(){
  int N;
  cin >> N;
  vector<int> A(N);
  for(int i = 0; i < N; ++i)
    cin >> A[i];
  vector<vector<int>> G(N);
  for(int i = 1; i < N; ++i){
    int u, v;
    cin >> u >> v;
    --u,--v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  vector<int> L(N+1,INF), R(N);
  dfs(G,A,L,R,0,-1);
  for(auto e : R)
    cout << e << "\n";
}
