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>
#define int long long
#define f first
#define s second
using namespace std;
const int N=2e5+9,inf=1e18;
int a[N],n,ans[N];
vector <int> g[N];
int dp[N];
void dfs(int v,int p){
	int pos=lower_bound(dp,dp+N,a[v])-dp;
	int tmp=dp[pos];
	dp[pos]=a[v];
	ans[v]=max(ans[p],pos);
	for(auto u:g[v]){
		if(u==p)continue;
		dfs(u,v);
	}
	dp[pos]=tmp;
}
int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    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;
    	g[u].push_back(v);
    	g[v].push_back(u);
    }
    fill(dp,dp+N,inf);
    dp[0]=0;
    dfs(1,0);
    for(int i=1;i<=n;i++)
    	cout << ans[i] << "\n";
}