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;

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

void dfs(int cur, int pre){
	int tmp = lower_bound(dp.begin(), dp.end(), a[cur]) - dp.begin();
	int par = dp[tmp];
	dp[tmp] = a[cur];
	ans[cur] = tmp;
	if(pre != -1) ans[cur] = max(ans[cur], ans[pre]);
	for(auto&& next : to[cur]){
		if(next == pre) continue;
		dfs(next, cur);
	}
	dp[tmp] = par;
}

int main()
{
	int n;
	cin >> n;
	to.resize(n);
	a = ans = vector<int>(n);
	dp = vector<int>(n, 1e9);
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int i = 0; i < n - 1; i++){
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		to[u].push_back(v);
		to[v].push_back(u);
	}
	dp[0] = -1e9;
	dfs(0, -1);
	for(int i = 0; i < n; i++) cout << ans[i] << endl;
	return 0;
}