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 N=2e5+20;

int n,a[N],u,v,ans[N];
vector <int> adj[N];
multiset <int> s;

void dfs(int node,int parent=0)
{
	s.insert(a[node]);
	int del=-1;
	auto it=s.lower_bound(a[node]);
	if(++it!=s.end()) del=*it,s.erase(it);
	ans[node]=s.size();
	
	for(auto child:adj[node])
		if(child!=parent) dfs(child,node);
	
	if(del!=-1) s.insert(del);
	s.erase(s.find(a[node]));
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=0;i<n-1;i++)
	{
		scanf("%d%d",&u,&v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}

	dfs(1);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}
