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
#pragma GCC optimize("Ofast")
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int n,num[200005],ans[200005];
vector<int> g[200005],dp(200005,2147483647);
void DFS(int v,int p){
	auto x=lower_bound(dp.begin(),dp.begin()+n,num[v]);
	int y=*x;
	*x=num[v];
	ans[v]=lower_bound(dp.begin(),dp.begin()+n,2147483647)-dp.begin();
	for(auto z:g[v])
		if(z!=p)
			DFS(z,v);
	*x=y;
}
int main(){
	int i,from,to;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		scanf("%d",&num[i]);
	for(i=0;i<n-1;i++){
		scanf("%d%d",&from,&to);
		g[from].emplace_back(to);
		g[to].emplace_back(from); 
	}
	DFS(1,-1);
	for(i=1;i<=n;i++)
		printf("%d\n",ans[i]);
}