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, a[200010], answer[200010];
vector<int> E[200010], vec{0};
void dfs(int u, int p) {
	int idx = -1, old = -1;
	if (a[u] > vec.back()) {
		vec.push_back(a[u]);
	} else {
		idx = lower_bound(vec.begin(), vec.end(), a[u]) - vec.begin();
		old = vec[idx];
		vec[idx] = a[u];
	}
	/*
	cout << u << " " << idx << '\n';
	for (int x : vec) {
		cout << x << " ";
	}
	cout << '\n';
	*/
	answer[u] = vec.size();
	for (int v : E[u]) {
		if (v != p) {
			dfs(v, u);
		}
	}
	if (idx == -1) {
		vec.pop_back();
	} else {
		vec[idx] = old;
	}
}
int main() {
	ios_base::sync_with_stdio(0); cin.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;
		E[u].push_back(v);
		E[v].push_back(u);
	}
	dfs(1, 1);
	for (int i = 1; i <= n; ++i) {
		cout << answer[i] - 1 << '\n';
	}
}