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 val[200010];
vector<int> edge[200010];
int ans[200010];

void dfs(int from, int par, vector<int>& dp) {
	if(dp.back() < val[from]) {
		ans[from] = dp.size();
		dp.push_back(val[from]);
		for(int to : edge[from]) {
			if(to == par) continue;
			dfs(to, from, dp);
		}
		dp.pop_back();
	} else {
		int idx = lower_bound(dp.begin(), dp.end(), val[from]) - dp.begin();
		int pre = dp[idx];
		ans[from] = dp.size() - 1;
		dp[idx] = val[from];
		for(int to : edge[from]) {
			if(to == par) continue;
			dfs(to, from, dp);
		}
		dp[idx] = pre;
	}
}

int main() {
	int n; cin >> n;
	for(int i = 0; i < n; ++i) cin >> val[i];
	for(int i = 0; i < n - 1; ++i) {
		int u, v; cin >> u >> v;
		--u, --v;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	vector<int> dp = {0};
	dfs(0, -1, dp);
	for(int i = 0; i < n; ++i) {
		cout << ans[i] << '\n';
	}
	return 0;
}