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+7;

int h[N],e[N*2],ne[N*2],idx;
int w[N];
int ans[N]; 
int low[N];

void add(int a,int b){
	e[idx] = b,ne[idx] = h[a] , h[a] = idx++;
}

void dfs(int s,int fa,int pos)
{
	int temp1=-1,temp2=0,k=0;
	if(low[pos] < w[s]) temp1 = low[pos+1] ,low[++pos] = w[s];  //做好记录
	else {
		int j = lower_bound(low+1,low+pos+1,w[s]) - low ;
		temp2 = low[j],k = j;
		low[j] = w[s];
	}
	ans[s] = pos;
	for(int i=h[s];~i;i=ne[i]){
		int j = e[i];
		if(j==fa) continue;
		dfs(j,s,pos);
	}
	if(temp1!=-1) low[pos] = temp1;			//回溯
	else low[k] = temp2;
}

int main()
{
	int n,a,b;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&w[i]);
	memset(h,-1,sizeof h);
	for(int i=1;i<n;i++){
		scanf("%d%d",&a,&b);
		add(a,b),add(b,a);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
	return 0;
}
