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;

const int N=500005,INF=0x3f3f3f3f;
int n,a[N],dp[N],len,ans[N];
int v[N],nxt[N],head[N];

void dfs(int x,int fa)
{
//	cout<<x<<' '<<fa<<endl;
	int L=1,R=n,id,ls;
	while(L<=R)
	{
		int mid=(L+R)>>1;
		if(dp[mid]<a[x]) L=mid+1;
		else id=mid,R=mid-1;
	}
	ls=dp[id],dp[id]=a[x];
	if(ls==INF) len++;
	ans[x]=len;
	for(int i=head[x];i>0;i=nxt[i])
	{
		if(v[i]==fa) continue;
		dfs(v[i],x);
	}
	dp[id]=ls;
	if(ls==INF) len--;
}

int main()
{
	memset(dp,0x3f,sizeof(dp));
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&v[i],&v[i+n]);
		nxt[i]=head[v[i+n]],head[v[i+n]]=i;
		nxt[i+n]=head[v[i]],head[v[i]]=i+n;
	}
	dfs(1,0);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}