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;
#define pb push_back
const int N=200050;
vector<int> E[N],lis;
int a[N],ans[N];
void DFS(int u,int p){
	bool pop=0;
	int las=0;
	int i=lower_bound(lis.begin(),lis.end(),a[u])-lis.begin();
	if(i==lis.size()){
		pop=1;
		lis.pb(a[u]);
	}else{
		las=lis[i];
		lis[i]=a[u];
	}
	ans[u]=lis.size();
	for(int v:E[u])if(v!=p)DFS(v,u);
	if(pop)lis.pop_back();
	else lis[i]=las;
}
int main(){
	int n;
	scanf("%i",&n);
	for(int i=1;i<=n;i++)scanf("%i",&a[i]);
	for(int i=1,u,v;i<n;i++)scanf("%i %i",&u,&v),E[u].pb(v),E[v].pb(u);
	DFS(1,0);
	for(int i=1;i<=n;i++)printf("%i\n",ans[i]);
	return 0;
}