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=200010;
int n,num_edge,a[N],f[N],ans[N],head[N];
struct Edge 
{
	int to; 
	int Next;
}edge[N<<1];
void add_edge(int from,int to) 
{
	edge[++num_edge].to=to;
	edge[num_edge].Next=head[from];
	head[from]=num_edge;
}
void dfs(int x,int fa) 
{
	int last;
	int l=1,r=ans[fa];
	if(fa==0)
	{
		f[1]=a[x];
		ans[1]=1;
	}
	else 
	{
		while(l<=r) 
		{
			int mid=(l+r)>>1;
			if(a[x]<=f[mid]) 
				r=mid-1;
			else 
				l=mid+1;
		}
		last=f[l];
		f[l]=a[x];
		ans[x]=ans[fa];
		if(l>ans[fa]) 
			ans[x]++;
	}
	for(int i=head[x];i;i=edge[i].Next) 
	{
		int y=edge[i].to;
		if(y!=fa) 
			dfs(y,x);
	}
	f[l] = last;
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n-1;i++) 
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add_edge(x,y);
		add_edge(y,x);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}
