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
//#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define int long long
using namespace std;
int n;
struct ss{
	int node,nxt;
}e[500005];
int tot,head[200005],a[200005],b[200005],l,f[200005],g[200005];
void add(int u,int v)
{
	e[++tot].nxt=head[u];
	e[tot].node=v;
	head[u]=tot;
}
void dfs(int x,int fa)
{
	int q=a[x],now=0;
	if(g[l]<a[x])g[++l]=a[x];
	else
	{
		now=lower_bound(g+1,g+l+1,q)-g;
		swap(g[now],q);
	}
	f[x]=l;
	for(int i=head[x];i;i=e[i].nxt)
	{
		if(e[i].node!=fa)dfs(e[i].node,x);
	}
	if(now)
	{
		swap(g[now],q);
	}
	else l--;
	
}
signed main()
{
	cin>>n;
	int u,v;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	for(int i=1;i<n;i++)
	{	
		cin>>u>>v;
		add(u,v);add(v,u);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		cout<<f[i]<<endl;
}