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 a[200005], lis[200005];

vector<int> g[200005], v;

void dfs(int u, int parant = -87) {
	bool type = 0;
	int i, boook;
	if(v.empty() || v.back() < a[u]) {
		v.push_back(a[u]);
	}
	else {
		i = lower_bound(v.begin(), v.end(), a[u]) - v.begin();
		boook = v[i];
		v[i] = a[u];
		type = 1;
	}
	lis[u] = v.size();
	for(int child:g[u]) if(child != parant)
		dfs(child, u);
	if(type) {
		v[i] = boook;
	} else {
		v.pop_back();
	}
}

int main() {	
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for(int i = 1; i < n; i++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1);
	for(int i = 1; i <= n; i++) {
		cout << lis[i] << endl;
	}

}