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>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

const int max_n = 202020;
int INF = 2e9;
vector<int> g[max_n];
vector<int> dp(max_n, INF);

vector<int> a;
vector<int> ans;

void dfs(int x, int p){
	auto itr = lower_bound(dp.begin(), dp.end(), a[x]);
	int i = itr - dp.begin();
	int from = *itr;
	int to = min(*itr, a[x]);
	*itr = to;
	
	ans[x] = lower_bound(dp.begin(), dp.end(), INF) - dp.begin();
	
	for(int y : g[x]){
		if(y == p) continue;
		dfs(y, x);
	}
	dp[i] = from;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;

	a.resize(n);
	rep(i, n) cin >> a[i];

	rep(i, n - 1){
		int u, v;
		cin >> u >> v;
		u--; v--;

		g[u].push_back(v);
		g[v].push_back(u);
	}

	ans.resize(n);
	dfs(0, -1);
	rep(i, n){
		cout << ans[i] << "\n";
	}

}
