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;
int a[200001], ans[200001];

vector<int> tmp;
vector<int> adj[200001];

void dfs(int u, int p)
{       
	int pre = -1;
	int pos = lower_bound(tmp.begin(), tmp.end(), a[u]) - tmp.begin();
	if (pos == tmp.size())
		tmp.push_back(a[u]);
	else
	{
		pre = tmp[pos];
		tmp[pos] = a[u];
	}
	ans[u] = tmp.size();
	for (int v : adj[u])
		if (v != p)	
			dfs(v, u);

	if (pre != -1) tmp[pos] = pre;
	else tmp.pop_back();
}

int main(int argc, char** argv)
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = 1; i <= n - 1; i++)
	{
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1, -1);
	for (int i = 1; i <= n; i++)
		cout << ans[i] << '\n';
}