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 pb emplace_back
using namespace std;
using ll = long long;
const int maxn = 300010;
vector<int> edge[maxn];
int n, a[maxn], res[maxn];
int dp[maxn];
void dfs(int now, int last = -1, int len = 0) {
	int pos = lower_bound(dp, dp+len, a[now]) - dp;
	int o = dp[pos];
	len = max(len, pos+1);
	//if (!len || a[now] > dp[len-1].back()) pos = len+1;
	//cerr << now << " : " << pos << '\n';
	dp[pos] = a[now];
	res[now] = len;
	for (int u : edge[now]) if (u != last) 
		dfs(u, now, len);
	dp[pos] = o;
}

signed main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for (int i = 1;i <= n;++i)
		cin >> a[i];
	for (int i = 1, a, b;i < n;++i) {
		cin >> a >> b;
		edge[a].pb(b);
		edge[b].pb(a);
	}
	dfs(1);
	for (int i = 1;i <= n;++i)
		cout << res[i] << '\n';
}

