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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using graph=vector<vector<int>>;

graph T;
vector<int> val;

vector<int> lis,ans;

void dfs(int u,int p){
	auto it=lower_bound(lis.begin(),lis.end(),val[u]);
	int pos,val_pre;
	if(it==lis.end()){
		pos=-1;
		lis.emplace_back(val[u]);
	}
	else{
		pos=it-lis.begin();
		val_pre=lis[pos];
		lis[pos]=val[u];
	}

	ans[u]=lis.size();
	for(int v:T[u]) if(v!=p) dfs(v,u);

	if(pos==-1) lis.pop_back();
	else        lis[pos]=val_pre;
}

int main(){
	int n; scanf("%d",&n);
	T.resize(n);
	val.resize(n);
	rep(u,n) scanf("%d",&val[u]);
	rep(i,n-1){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		T[u].emplace_back(v);
		T[v].emplace_back(u);
	}

	ans.resize(n);
	dfs(0,-1);

	rep(u,n) printf("%d\n",ans[u]);

	return 0;
}
