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>
#define ll long long
#define vi vector<int>
#define pb push_back
using namespace std;
const int N=2e5+5;
vi v[N],e;
int n,m,x[N],ans[N];
void dfs(int now,int fa)
{
	int rem=-1,num;
	if(e.empty()||x[now]>e.back())
		e.pb(x[now]);
	else{
		num=lower_bound(e.begin(),e.end(),x[now])-e.begin();
		rem=e[num];
		e[num]=x[now];
	}
	ans[now]=e.size();
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fa)continue;
		dfs(to,now);
	}
	if(rem==-1)
		e.pop_back();
	else
		e[num]=rem;
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>x[i];
	for(int i=1;i<n;i++){
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
		v[b].pb(a);
	}
	dfs(0,-1);
	for(int i=0;i<n;i++)cout<<ans[i]<<"\n";
}