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;

int n;
int a[201000];
int ans[201000];
int dis[201000];
vector<int> e[210000];

void dfs(int x,int y,int l)
{   int pos = 210000,tmp;
	if(a[x] > dis[l]) dis[++l] = a[x];
	else{
		pos = lower_bound(dis+1,dis+l+1,a[x])-dis;
		tmp = dis[pos];
		dis[pos] = a[x];
	}
	ans[x] = max(ans[x],l);
    for(int i = 0;i < e[x].size();i++)
    {
    	if(e[x][i] != y)
    	{
    		dfs(e[x][i],x,l);
		}
	}
    if(pos != 210000)	
	dis[pos] = tmp;
}
int main(){
   cin>>n;
   
   for(int i = 1;i <= n;i++)
     cin>>a[i];
   for(int i = 1;i <= n-1;i++)
   {
   	int u,v;
   	scanf("%d%d",&u,&v);
    e[u].push_back(v);
    e[v].push_back(u);
   }
   dis[1] = a[1];
   dfs(1,0,1);
   for(int i = 1;i <=n;i++)
    cout<<ans[i]<<endl;
}