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
import java.util.*;
import java.io.PrintWriter;
public class Main{
	static ArrayList<Integer>[] list;
	static int[] a;
	static int[] dp;
	static int[] ans;
	static int max;
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		list=new ArrayList[n];
		a=new int[n];
		dp=new int[n+1];
		ans=new int[n];
		max=0;
		Arrays.fill(dp,1000000001);
		dp[0]=0;
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
			list[i]=new ArrayList<>();
		}
		for(int i=1; i<n; i++){
			int u=sc.nextInt()-1;
			int v=sc.nextInt()-1;
			list[u].add(v);
			list[v].add(u);
		}
		dfs(0,0);
		PrintWriter pw=new PrintWriter(System.out);
		for(int i=0; i<n; i++){
			pw.println(ans[i]);
		}
		pw.flush();
	}
	static void dfs(int now,int from){
		int idx=Arrays.binarySearch(dp,a[now]);
		if(idx<0){
			idx=-idx-1;
		}
		int tmp=idx;
		int val=dp[idx];
		dp[idx]=a[now];
		int pp=max;
		max=Math.max(max,idx);
		ans[now]=max;
		
		for(int i:list[now]){
			if(i!=from){
				dfs(i,now);
			}
		}
		max=pp;
		dp[tmp]=val;
	}
}