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>
typedef long long int ll;

using namespace std;
const int N=2*1e5+5;
vector<vector<int>>g(N);
vector<int>h(N);
multiset<int>s;
int ans[N];

void dfs(int node,int p=0)
{
	
	s.insert(h[node]);
	int prev=-1;
	auto it=s.lower_bound(h[node]);
	if(++it!=s.end())
	{
		prev=*it;
		s.erase(it);
	}
	ans[node]=s.size();
	for(int v:g[node])
	{
		if(v!=p)
		dfs(v,node);
	}
	   
	   
	if(prev!=-1)
	s.insert(prev);
	
	s.erase(s.find(h[node]));
	
	
}
int main() {
	// your code goes here
	int t;
	t=1;
	///solve();
	while(t--)
	{
		int n;
		cin>>n;
		for(int i=1;i<=n;i++)
		{
			cin>>h[i];
			
		}
		
		for(int i=1;i<n;i++)
		{
			int u,v;
			cin>>u>>v;
			g[u].push_back(v);
			g[v].push_back(u);
		}
		dfs(1);
	//	cout<<n<<"\n";
		for(int i=1;i<=n;i++)
		cout<<ans[i]<<"\n";
		
		
			
	
		
	}
	return 0;
}