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>
#define Int int64_t

using namespace std;

int main() {
	constexpr int INF = 1e9 + 10;
	int N;
	cin >> N;
	vector<int> a(N);
	for (int i = 0; i < N; ++i) { cin >> a[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, 0);
	vector<bool> used(N, false);
	used[0] = true;
	vector<int> dp(N, INF);
	dp[0] = a[0];
	function<void(int)> dfs = [&](int u) -> void {
		ans[u] = lower_bound(dp.begin(), dp.end(), INF) - dp.begin();
		for (auto v : g[u]) {
			if (used[v]) { continue; }
			used[v] = true;
			int idx = lower_bound(dp.begin(), dp.end(), a[v]) - dp.begin();
			int tmp = dp[idx];
			dp[idx] = a[v];
			dfs(v);
			dp[idx] = tmp;
		}
		return;
	};

	dfs(0);
	for (auto v : ans) { cout << v << endl; }

	return 0;
}
