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;
typedef pair<int,int> P;
const int MAX_N = 200005;
const int INF = (1e9)+5;
bool visited[MAX_N];
vector<int> G[MAX_N];
int a[MAX_N],dp[MAX_N],ans[MAX_N];
int N;

void init()
{
	rep(i,MAX_N)
	{
		dp[i] = INF;
	}
}

void dfs(int x)
{
	if (visited[x]) return;
	visited[x] = true;
	auto it = lower_bound(dp, dp+N, a[x]);
	int idx = it - dp;
	int v = *it;
	*it = a[x];
	ans[x] = lower_bound(dp, dp+N, INF) - dp;
	rep(i,G[x].size())
	{
		dfs(G[x][i]);
	}
	dp[idx] = v;
	return;
}

int main()
{
	init();
	cin >> 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);
	}
	vector<int> dp;
	dp.push_back(a[0]);
	dfs(0);
	rep(i,N)
	{
		cout << ans[i] << endl;
	}
	return 0;
}