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;
int N, A[200010]; 
vector<int> E[200010]; 
int ans[200010]; 
int LIS[200010];
pair<int, int> backup[200010];   
int INF = 1145141919; 
void dfs(int v, int p){
	int a = lower_bound(LIS, LIS + N, A[v]) - LIS; 
	backup[v] = make_pair(a, LIS[a]); 
	LIS[a] = A[v]; 
	int b = lower_bound(LIS, LIS + N, INF) - LIS; 
	ans[v] = b; 
	for(auto i : E[v]){
		if(i == p) continue; 
		dfs(i, v); 
	}
	LIS[backup[v].first] = backup[v].second;  
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0); 
	cin >> N;  
	for(int i = 0; i < N; i++) cin >> A[i]; 
	for(int i = 0; i < N-1; i++){
		int u, v; cin >> u >> v; 
		u--; v--; 
		E[u].push_back(v); E[v].push_back(u); 
	}
	fill(LIS, LIS + 200005, INF); 
	dfs(0, -1); 
	for(int i = 0; i < N; i++) cout << ans[i] << "\n"; 
}