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;
vector<int>ele;
int a[N+100],ans[N+100];
vector<int>g[N+100];

void dfs(int v,int par,int cur)
{
	auto in = lower_bound(ele.begin(),ele.end(),a[v])-ele.begin();
	int pre = ele[in];
	if(in==cur){
		cur++;
	}
	ele[in] = a[v];
	ans[v] = cur;
	for(auto it:g[v])
	{
		if(it!=par){
			dfs(it,v,cur);
		}
	}
	ele[in] = pre;
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",a+i);
	}
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for(int i=1;i<=n+100;i++){
		ele.push_back(1e9+1);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++){
		printf("%d\n",ans[i]);
	}
}
