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=2e5+5;
struct Edge{int next,to;}edge[N*2];
int fi[N],se,a[N],ans[N],b[N],s;
inline void add_edge(int u,int v){edge[++se]={fi[u],v},fi[u]=se;}
inline void add(int u,int v){add_edge(u,v),add_edge(v,u);}
void dfs(int x,int fa){
	int w=lower_bound(b,b+s,a[x])-b,w1=b[w];
	b[w]=a[x];ans[x]=s=max(s,w+1);
	for(int i=fi[x];i;i=edge[i].next){
		int v=edge[i].to;if(v==fa)continue;
		dfs(v,x);s=ans[x];
	}
	b[w]=w1;
}
int main(){
	int n,u,v;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",a+i);
	for(int i=1;i<n;i++)scanf("%d%d",&u,&v),add(u,v);
	dfs(1,0);
	for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
	return 0;
}