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;
typedef long long ll;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const int N=2e5+5;
const int mod=1e9+7;

vector<int> v[N];
int val[N],cnt;
int dp[N],ans[N];

void dfs(int u,int par=-1){
	int id=lower_bound(dp,dp+N,val[u])-dp;
	int pre=dp[id];
	dp[id]=val[u];
	ans[u]=id;
	if(par!=-1) ans[u]=max(ans[u],ans[par]);
	for(int x:v[u]){
		if(x==par) continue;
		dfs(x,u);
	}
	dp[id]=pre;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;++i) cin>>val[i];
	for(int i=0;i<n-1;++i){
		int x,y;
		cin>>x>>y;
		--x;
		--y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	for(int i=0;i<N;++i) dp[i]=INT_MAX;
	dp[0]=-INT_MAX;
	dfs(0);
	for(int i=0;i<n;++i) cout<<ans[i]<<endl;
	return 0;
}
