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>

using namespace std;

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

const int N = 2e5;
int a[N], r[N], b[N+1];
vector<int> e[N];

int find(int v) {
  int l = 0, r = N;
  while(l + 1 < r) {
    int m = (r + l) / 2;
    if(b[m] < v) l = m;
    else r = m;
  }
  if(v <= b[l]) return l;
  else return r;
}

int dfs(int c, int p) {
  int x = find(a[c]), y = b[x];
  b[x] = a[c];
  r[c] = find(1e9+1);
  for(auto i : e[c]) {
    if(i != p) dfs(i, c);
  }
  b[x] = y;
  return 0;
}

int main() {
  int n;
  cin >> n;
  rep(i,n) cin >> a[i];
  rep(i,n-1) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    e[u].push_back(v);
    e[v].push_back(u);
  }

  rep(i,N+1) b[i] = 1e9+1;
  dfs(0, -1);

  rep(i,n) cout << r[i] << endl;
  return 0;
}