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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>

using namespace std; using ll=long long; const int INF=1e9;
typedef pair<int,int> P;
const int mx = 200000;

vector<int> v[mx]; int an[mx], a[mx]; int n; int pp[mx];

void dfs(int from, int now, int pc) {
  //cout<<from<<" "<<now<<" "<<pc<<" ";
  auto it = lower_bound(pp, pp+mx, a[now]);
  int pc2 = max(pc, (int)(it-pp));
  //cout<<pc2<<" ";
  int bf = *it;
  //cout<<bf<<endl;
  *it = a[now];
  an[now] = pc2;
  
  for(int x:v[now]) {
    if (x == from) continue;
    dfs(now, x, pc2);
  }

  *it = bf;
}

int main() {
  cin>>n;
  for(int i=0;i<n;i++) cin>>a[i];

  for(int i=0;i<n-1;i++) {
    int s,b; cin>>s>>b; s--; b--;
    v[s].push_back(b); v[b].push_back(s);
  }

  fill(pp, pp+mx, INF);

  dfs(-1, 0, 0);

  for(int i=0;i<n;i++) {
    cout<<an[i]+1<<endl;
  } 
}