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;
 
#define debug(P) cout<<#P<<" = "<<P<<endl;
 
typedef long long ll;


const int MAXN = 200001;
int n;
int a[MAXN];
int ans[MAXN];

vector<int> d;

vector<int> adj[MAXN];

void dfs(int x, int p){
	int pos = lower_bound(d.begin(), d.end(), a[x]) - d.begin();

	bool pushed = 0;

	if(pos == d.size())d.push_back(a[x]), pushed = 1;

	ans[x] = d.size();

	int svi = pos, svv = d[pos];

	if(!pushed)d[pos] = a[x];

	for(int c : adj[x]){
		if(c!=p){
			dfs(c, x);
		}
	}

	if(!pushed)d[svi] = svv;
	else d.pop_back();
}
 
int main() {
	//freopen("input.txt","r",stdin);
	cin>>n;

	for(int i= 1; i <= n; ++i)cin>>a[i];

	for(int i= 1; i < n; ++i){
		int u, v;
		cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}

	d.clear();

 	dfs(1, 0);

 	for(int i= 1; i <= n; ++i)cout<<ans[i]<<'\n';
}