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<vector>
using namespace std;
int N;
int A[2<<17];
vector<int>G[2<<17];
int dp[2<<17];
int ans[2<<17];
void dfs(int u,int p,int R)
{
	int l=0,r=R;
	while(r-l>1)
	{
		int m=l+r>>1;
		if(dp[m]<A[u])l=m;
		else r=m;
	}
	int pre=dp[r];
	dp[r]=A[u];
	ans[u]=R==r?R++:R-1;
	for(int v:G[u])if(v!=p)dfs(v,u,R);
	dp[r]=pre;
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)cin>>A[i];
	for(int i=1;i<N;i++)
	{
		int u,v;cin>>u>>v;u--,v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for(int i=1;i<=N;i++)dp[i]=2e9;
	dfs(0,-1,1);
	for(int i=0;i<N;i++)cout<<ans[i]<<endl;
}
