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;
#define int long long
vector<int> dp, arr;
vector<int> adj[200001];
int ans[200001];
void dfs(int i, int p) {
	auto it = lower_bound(dp.begin(), dp.end(), arr[i]);
	int rec = *it;
	*it = arr[i];
	ans[i] = lower_bound(dp.begin(), dp.end(), INT_MAX) - dp.begin() - 1;
	for(int a: adj[i]) {
		if(a != p)
			dfs(a, i);
	}
	*it = rec;
}
int32_t main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n;
	cin >> n;
	dp = arr = vector<int>(n+1);
	fill(dp.begin(), dp.end(), INT_MAX);
	dp[0] = 0;
	for(int i=1; i<=n; i++) {
		cin >> arr[i];
	}
	for(int i=0; i<n-1; i++) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1, 0);
	for(int i=1; i<=n; i++) {
		cout << ans[i] << '\n';
	}
}