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 main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;

	vector<int> a(n);
	for(int &i : a) {
		cin >> i;
	}

	vector<vector<int> > G(n);

	for(int i = 0; i < n - 1; ++i) {
		int u, v;
		cin >> u >> v;
		--u; --v;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}

	vector<int> ans(n);

	int mx = *max_element(a.begin(), a.end());

	vector<int> dp(n, mx + 1);

	function<void(int, int)> dfs = [&] (int u, int f) {
		int p = lower_bound(dp.begin(), dp.end(), a[u]) - dp.begin();
		ans[u] = max(ans[u], p + 1);
		int tmp = dp[p];
		dp[p] = a[u];
		for(int v : G[u]) {
			if(v != f) {
				ans[v] = max(ans[v], ans[u]);
				dfs(v, u);
			}
		}
		dp[p] = tmp;
	};

	dfs(0, -1);

	for(int i : ans) {
		cout << i << '\n';
	}

	return 0;
}
