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;
#define int long long
const int MOD = 1000000007;

vector<vector<int> > g;
vector<int> res;
vector<int> A;
vector<int> X;
int N;
int INF = (int)1 << 62;
void dfs(int a, int p = -1) {
	int pos = lower_bound(X.begin(), X.begin() + N, A[a]) - X.begin();
	int b = X[pos];
	X[pos] = A[a];
	res[a] = lower_bound(X.begin(), X.begin() + N, INF) - X.begin();
	for (int i : g[a]) {
		if (i != p) {
			dfs(i, a);
		}
	}
	X[pos] = b;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> N;
	g.resize(N);
	res.resize(N);
	X.resize(N, INF);
	A.resize(N);
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}

	for (int i = 0; i < N - 1; i++) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		g[a].push_back(b);
		g[b].push_back(a);
	}

	dfs(0);
	for (int i = 0; i < N; i++) {
		cout << res[i] << endl;
	}
}