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;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

int n;
int a[300000];
vector<int>E[300000];
int dp[300000];
int ans[300000];

void dfs(int v,int p){
	int id=lower_bound(dp,dp+300000,a[v])-dp;
	int memo=dp[id];
	dp[id]=a[v];
	ans[v]=lower_bound(dp,dp+300000,INF)-dp;
	for(int u:E[v]){
		if(u==p)continue;
		dfs(u,v);
	}
	dp[id]=memo;
}
int main(){
	cin>>n;
	rep(i,n)scanf("%d",&a[i]);
	rep(i,n-1){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
		E[b].push_back(a);
	}
	memset(dp,0x3f,sizeof(dp));
	dfs(0,-1);
	rep(i,n){
		printf("%d\n",ans[i]);
	}
}