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 <iostream>
#include <vector>
#include <stack>
#include <algorithm>

using namespace std;

const int INF = 2 * 1000 * 1000 * 1000;

void solve(int N, const vector<int> &a, const vector<vector<int>>& g, int p, int x, vector<int> &lis, vector<int>& ans)
{
	auto idx = lower_bound(lis.begin(), lis.end(), a[x]) - lis.begin();
	auto val = lis[idx];
	lis[idx] = a[x];
	ans[x] = lower_bound(lis.begin(), lis.end(), INF) - lis.begin();
	for (auto y : g[x]) {
		if (y != p) {
			solve(N, a, g, x, y, lis, ans);
		}
	}
	lis[idx] = val;
}

int main()
{
	int N;
	cin >> N;
	vector<int> a(N);
	for (auto& x : a) {
		cin >> x;
	}
	vector<vector<int>> g(N);
	for (int i = 0; i < N - 1; ++i) {
		int u, v;
		cin >> u >> v;
		--u;
		--v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	vector<int> lis(N, INF);
	vector<int> ans(N);
	solve(N, a, g, -1, 0, lis, ans);
	for (auto x : ans) {
		cout << x << endl;
	}
}
