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"
using namespace std;

int N;
vector<int> A;
vector<vector<int> > V;
vector<int> ANS;
vector<int> NOW;

void DFS(int P, int Q) {
	ANS[P] = NOW.size();
	for (int NP : V[P]) {
		if (NP == Q) continue;
		if (NOW[NOW.size() - 1] < A[NP]) {
			NOW.push_back(A[NP]);
			DFS(NP, P);
			NOW.pop_back();
		}
		else {
			int X = lower_bound(NOW.begin(), NOW.end(), A[NP]) - NOW.begin();
			int Y = NOW[X];
			NOW[X] = A[NP];
			DFS(NP, P);
			NOW[X] = Y;
		}
	}
}

int main() {
	cin >> N;
	A.resize(N);
	V.resize(N);
	ANS.resize(N);
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < N - 1; i++) {
		int S, T;
		cin >> S >> T;
		S--, T--;
		V[S].push_back(T);
		V[T].push_back(S);
	}
	NOW.push_back(A[0]);
	DFS(0, -1);
	for (int i = 0; i < N; i++) {
		cout << ANS[i] << endl;
	}
}