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 Maxn 200010

int n, a[Maxn], x, y, Ans[Maxn];
multiset<int >Rep; 
vector<int >Edge[Maxn];

inline void dfs(int u, int pre)
{
	Rep.insert(a[u]);
	auto cur = Rep.lower_bound(a[u]);
	int Keep = -1;
	if (++cur != Rep.end()) Keep = *cur, Rep.erase(cur);
	Ans[u] = Rep.size();
	
	for (auto v: Edge[u]) if (v != pre) dfs(v, u);
  
	if (Keep != -1) Rep.insert(Keep);
	Rep.erase(Rep.find(a[u]));
}

int main()
{
	scanf("%d", &n);
	for (int i=1; i<=n; i++) scanf("%d", &a[i]);
	for (int i=1; i<n; i++)
	{
		scanf("%d%d", &x, &y);
		Edge[x].push_back(y);
		Edge[y].push_back(x);
	}
	
	dfs(1, 0);
	for (int i=1; i<=n; i++) printf("%d\n", Ans[i]);
}