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>
#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
int n,a[200004];
vector<int>g[200004];
int ans[200004],len;
int pos[200004],lst[200004];
int ret[200004];
void dfs(int x,int par){
	if(ans[len]<a[x])pos[x]=len+1,lst[x]=0,ans[++len]=a[x];
	else{
		pos[x]=lower_bound(ans+1,ans+1+len,a[x])-ans;
		lst[x]=ans[pos[x]];
		ans[pos[x]]=a[x];
	}
	ret[x]=len;
	for(int i=0;i<g[x].size();i++)if(g[x][i]!=par)
		dfs(g[x][i],x);
	if(lst[x]==0)len--;
	else ans[pos[x]]=lst[x];
}
int main(){
	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);
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++)printf("%d\n",ret[i]);
}