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;
const int INF = (int)1e10;

vector<vector<int>> e;
vector<int> dp, o, a;

void dfs(int v, int p = -1) {
  auto it = lower_bound(dp.begin(), dp.end(), a[v]);
  int prev = *it;
  *it = a[v];
  for(auto x : e[v])
    if(x != p) dfs(x, v);
  o[v] = distance(dp.begin(), lower_bound(dp.begin(), dp.end(), INF)) - 1;
  *it = prev;
}

int main() {
  int n;
  scanf("%d", &n);
  e = vector<vector<int>>(n);
  dp = vector<int>(n + 1, INF);
  o = vector<int>(n);
  a = vector<int>(n);
  dp[0] = -INF;
  for(int i = 0; i < n; i++) scanf("%d", &a[i]);
  for(int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    e[--u].push_back(--v);
    e[v].push_back(u);
  }
  dfs(0);
  for(int i = 0; i < n; i++) printf("%d\n", o[i]);
}