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;
int n,a[202020],ans[202020],nw[202020];
vector<int>v[202020];
void dfs(int x,int y,int nww)
{
	int vl=a[x],p=lower_bound(nw,nw+n,vl)-nw,pp=nw[p];
	nw[p]=vl;
	ans[x]=max(nww,p+1);
	for(int i=0;i<v[x].size();i++)
	{
		int z=v[x][i];
		if(z!=y)
		{
			dfs(z,x,ans[x]);
		}
	}
	nw[p]=pp;
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	for(int i=0;i<n-1;i++)
	{
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for(int i=0;i<n;i++)
	{
		nw[i]=1e9+100;
	}
	dfs(0,-1,0);
	for(int i=0;i<n;i++)
	{
		cout<<ans[i]<<endl;
	}
	return 0;
}