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 <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;

int N, A[200200], R[200200];
vector<int> G[200200], L;

void go(int x, int l)
{
	auto i = lower_bound(L.begin(), L.end(), A[x]) - L.begin();
	int v = -1;

	if (i == L.size()) L.push_back(A[x]);
	else v = L[i], L[i] = A[x];

	R[x] = L.size();

	for (auto &y : G[x]) if (y != l) go(y, x);

	if (v == -1) L.pop_back();
	else L[i] = v;
}

int main()
{
	scanf ("%d", &N);
	for (int i = 1; i <= N; i++) scanf ("%d", &A[i]);
	for (int i = 1; i < N; i++){
		int x, y; scanf ("%d %d", &x, &y);
		G[x].push_back(y);
		G[y].push_back(x);
	}

	go(1, 0);

	for (int i = 1; i <= N; i++) printf ("%d\n", R[i]);

	return 0;
}