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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int MR=2e5+10;
typedef long long LL;

int a[MR],p[MR],ans[MR],plen;
bool vis[MR];
vector<int> d[MR];

void dfs(int idx){
	int rawplen=plen,rawlow,rawp;
	int* low=lower_bound (p, p+plen, a[idx]); 
	if(low==p+plen) {
		p[plen]=a[idx];
		plen++;
	}
	else{
		rawlow=low-p;
		rawp=p[rawlow];
		p[rawlow]=a[idx];
	}
	ans[idx]=plen;
	vis[idx]=true;
	int len=d[idx].size();
	for(int j=0;j<len;j++){
		if(vis[d[idx][j]]) continue;
		dfs(d[idx][j]);
	}


	if(plen!=rawplen) plen--;
	else p[rawlow]=rawp;
}

int main() {
	int n,u,v;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i=1;i<n;i++){
		cin>>u>>v;
		d[u].push_back(v);
		d[v].push_back(u);
	}
	plen=0;
	dfs(1);
	for(int i=1;i<=n;i++){
		printf("%d\n",ans[i]);
	}
    return 0;
}