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 <string.h>
#define int long long
using namespace std;
int N;
int pos[200100];
int bef[200100];
int arr[200100];
int a[200100];
vector<int>li[200100];
vector<int>p;
void dfs(int n,int pa)
{
	int i;
	pos[n] = lower_bound(p.begin(), p.end(), a[n])-p.begin();
	if (pos[n] < p.size())
	{
		bef[n] = p[pos[n]];
		p[pos[n]] = a[n];
	}
	else
		p.push_back(a[n]);
	arr[n] = p.size();
	for (i = 0; i < li[n].size(); i++)
	{
		if (li[n][i] == pa)
			continue;
		dfs(li[n][i], n);
	}
	if (bef[n])
	{
		p[pos[n]] = bef[n];
	}
	else
		p.pop_back();
}
signed main()
{
	ios_base::sync_with_stdio(false);
	cin >> N;
	int i;
	for (i = 1; i <= N; i++)
		cin >> a[i];
	for (i = 1; i < N; i++)
	{
		int a, b;
		cin >> a >> b;
		li[a].push_back(b);
		li[b].push_back(a);
	}
	dfs(1, 0);
	for (i = 1; i <= N; i++)
	{
		cout << arr[i]<<'\n';
	}
}