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
//love yjl forever
#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
const int INF=0x3f3f3f3f,N=2e5+5;
int n;
int a[N];
int ans[N];
bool vis[N];
int b[N];
int tot=1;
vector<int> v[N],v1;
void dfs(int x)
{
	ans[x]=tot;
	vis[x]=1;
	for(int i=0;i<v[x].size();i++)
	{
		int u=v[x][i];
		if(vis[u]) continue;
		
		int pos=lower_bound(b+1,b+tot+1,a[u])-b;
		if(pos>tot)
		{
			tot++;
			b[tot]=a[u];
			dfs(u);
			b[tot]=0;
			tot--;
		}
		else 
		{
			int tmp=b[pos];
			b[pos]=a[u];
			dfs(u);
			b[pos]=tmp;
		}
	}
}
int main()
{
/*	freopen("","r",stdin);
	freopen("","w",stdout);*/
	int i;
	cin>>n;
	for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		v[x].pb(y);
		v[y].pb(x);
	}
	b[1]=a[1];
	dfs(1);
	
	for(i=1;i<=n;i++)
		cout<<ans[i]<<endl;
}