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 <algorithm>
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
#define repl(i, l, r) for (ll i = l; i < r; i++)
#define rep(i, n) repl(i, 0, n)
using ll = long long;
vector<int> lis(200005, 1e9), G[200005];
int ans[200005], a[200005];
void DFS(const int& v, stack<pair<int, int>>& s) {
	int k = lower_bound(lis.begin(), lis.end(), a[v]) - lis.begin();
	s.push({k, lis[k]});
	lis[k] = a[v];
	ans[v] = lower_bound(lis.begin(), lis.end(), 1e9) - lis.begin();
	for (auto nv : G[v]) {
		if (ans[nv] == 0) DFS(nv, s);
	}
	int p = s.top().first, q = s.top().second;
	s.pop();
	lis[p] = q;
	return;
}
int main() {
	int n;
	scanf("%d", &n);
	rep(i, n) scanf("%d", &a[i]);
	rep(i, n - 1) {
		int u, v;
		scanf("%d%d", &u, &v);
		u--, v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	stack<pair<int, int>> s;
	DFS(0, s);
	rep(i, n) printf("%d\n", ans[i]);
	return 0;
}
