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;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int n,a[200005],ans[200005];
vector<int> g[200005];
multiset<int> s;
void dfs(int x,int y){
	s.insert(a[x]);
	multiset<int>::iterator it=s.lower_bound(a[x]);
	int z=0;
	if(++it!=s.end()){
		z=*it;
		s.erase(s.find(z));
	}
	ans[x]=s.size();
	for(int i=0;i<g[x].size();i++){
		int ch=g[x][i];
		if(ch==y)continue;
		dfs(ch,x);
	}
	if(z)s.insert(z);
	s.erase(s.find(a[x]));
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		g[x].push_back(y);
		g[y].push_back(x);
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}