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 <set>
#include <vector>
#include <iostream>

using namespace std;
struct _ { ios_base::Init i; _() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); } } _;

int n;
set<int> s;
vector<int> a, ans;
vector<vector<int>> gr;

void dfs(int curr, int par) {
	auto it = s.lower_bound(a[curr]);
	int temp = 0;
	if (it != s.end()) temp = *it, s.erase(it);
	s.insert(a[curr]);

	ans[curr] = s.size();

	for (auto child : gr[curr]) if (child != par) dfs(child, curr);

	s.erase(a[curr]);
	if (temp) s.insert(temp);
	return;
}

int main() {

	cin >> n;
	a.resize(n + 1, 0); for (int i = 1; i <= n; i++) cin >> a[i];
	gr.resize(n + 1);
	for (int i = 1; i <= n - 1; i++) {
		int u, v; cin >> u >> v;
		gr[u].push_back(v);
		gr[v].push_back(u);
	}

	ans.resize(n + 1, 0);
	dfs(1, 1);

	for (int i = 1; i <= n; i++) cout << ans[i] << "\n";

	return 0;
}