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 <stdio.h>
#define maxn 200010
int a[maxn],head[maxn],tot,n,d[maxn],ans,mx[maxn];
struct node{
	int to,next;
}e[maxn<<1];
void add_edge(int u,int v){
	tot++,e[tot].to=v,e[tot].next=head[u],head[u]=tot;
}
void dfs(int u,int fa){
	int b,v,save,flag,l,r,mid;;
	for(b=head[u];b;b=e[b].next){
		v=e[b].to;
		if(v==fa)continue;
		save=ans,flag=0;
		if(d[ans]<a[v])d[++ans]=a[v];
		else{
			l=0,r=ans;
			while(l+1<r){
				mid=(l+r)/2;
				if(d[mid]<a[v])l=mid;
				else r=mid;
			}
			flag=d[r],d[r]=a[v];
		}
		mx[v]=ans;
		dfs(v,u);
		ans=save;
		if(flag)d[r]=flag;
	}
}
void init(){
	int i,u,v;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<n;i++){
		scanf("%d%d",&u,&v);
		add_edge(u,v),add_edge(v,u);
	}
}
int main(){
	init();
	ans=1,d[1]=a[1],mx[1]=1;
	dfs(1,0);
	for(int i=1;i<=n;i++)printf("%d\n",mx[i]);
	return 0;
}