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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N, A[1 << 18];
int X[1 << 18], Y[1 << 18], Answer[1 << 18];
vector<int> G[1 << 18];

// LIS Array
int SZ;
int B[1 << 18];
pair<int, int> V[1 << 18];

void dfs(int pos, int dep, int pre) {
	int pos1 = lower_bound(B, B + SZ, A[pos]) - B;
	V[dep] = make_pair(pos1, B[pos1]);
	B[pos1] = A[pos];
	if (pos1 == SZ) SZ++;
	Answer[pos] = SZ;

	for (int i : G[pos]) {
		if (i == pre) continue;
		dfs(i, dep + 1, pos);
	}

	if (V[dep].second == 0) SZ--;
	B[V[dep].first] = V[dep].second;
	V[dep] = make_pair(0, 0);
}

int main() {
	cin >> N;
	for (int i = 1; i <= N; i++) cin >> A[i];
	for (int i = 1; i <= N - 1; i++) {
		cin >> X[i] >> Y[i];
		G[X[i]].push_back(Y[i]);
		G[Y[i]].push_back(X[i]);
	}
	dfs(1, 0, -1);

	for (int i = 1; i <= N; i++) cout << Answer[i] << endl;
	return 0;
}