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;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

int ret[200020];
int dp[200020];
int a[200020];
vector<int> d[200020];
int n;

void dfs(int x, int p, int l){
  auto lo = lower_bound(dp, dp + l, a[x]);
  int b = *lo;
  if(lo == dp + l){
    ret[x] = l + 1;
  }else{
    ret[x] = l;
  }
  *lo = a[x];
  for(auto y: d[x]) if(y != p){
    dfs(y, x, ret[x]);
  }
  *lo = b;
}

int main(){
  cin >> n;
  REP(i, n) cin >> a[i];
  REP(i, n - 1){
    int x, y;
    cin >> x >> y;
    --x, --y;
    d[x].push_back(y);
    d[y].push_back(x);
  }
  REP(i, n + 1){
    dp[i] = 1 << 30;
  }
  dfs(0, -1, 0);
  REP(i, n){
    cout << ret[i] << endl;
  }
  return 0;
}
