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;
vector<vector<int>> g;
vector<int> a, dp, ans;

void dfs(int x, int p = -1){
	auto it = lower_bound(dp.begin() + 1, dp.end(), a[x]) - dp.begin();
	int old = dp[it];
	dp[it] = a[x];
	ans[x] = it;
	if(p != -1)
		ans[x] = max(ans[x], ans[p]);
	for(int y : g[x]){
		if(y != p)
			dfs(y, x);
	}
	dp[it] = old;
}

int main(void){
	cin >> n;
	a.resize(n + 1);
	dp.resize(n + 1);
	ans.resize(n + 1);
	g.resize(n + 1);
	for(int i = 1 ; i <= n ; i++) cin >> a[i];
	for(int i = 1 ; i < n ; i++){
		int x, y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i = 1 ; i <= n ; i++){
		dp[i] = INT_MAX;
	}
	dfs(1);
	for(int i = 1 ; i <= n ; i++){
		cout << ans[i] << endl;
	}
	return 0;
}