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;
vector<int>v[200003],dp;
int n,a[200003],ans[200003];
void dfs(int x,int p){
	int idx=lower_bound(dp.begin(),dp.end(),a[x])-dp.begin(),tmp=dp[idx];
	dp[idx]=a[x];ans[x]=lower_bound(dp.begin(),dp.end(),INT_MAX)-dp.begin();
	for(int i=0;i<v[x].size();i++)
		if(v[x][i]!=p)
			dfs(v[x][i],x);
	dp[idx]=tmp;
}
int main(){
	scanf("%d",&n);dp=vector<int>(n,INT_MAX);
	for(int i=0;i<n;i++)scanf("%d",a+i);
	for(int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);x--;y--;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(0,-1);
	for(int i=0;i<n;i++)cout<<ans[i]<<endl;
}