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>
using namespace std;
const int INF = 1001001001;
const int MAX = 200005;

int a[MAX], ans[MAX];
vector<vector<int>> g;

void dfs(int v, int parent, vector<int> &dp) {
	auto itr = lower_bound(dp.begin(), dp.end(), a[v]);
	int tmp = *itr;
	*itr = a[v];
	ans[v] = lower_bound(dp.begin(), dp.end(), INF) - dp.begin();
	for (int nv : g[v]) {
		if (nv == parent) continue;
		dfs(nv, v, dp);
	}
	*itr = tmp;
}

int main() {
	int n;
	cin >> n;
	g.resize(n);
	for (int i = 0; i < n; i++) {
		scanf("%d", a+i);
	}
	for (int i = 0; i < n-1; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		u--, v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	vector<int> dp(n, INF);
	dfs(0, -1, dp);
	for (int i = 0; i < n; i++) {
		cout << ans[i] << endl;
	}
	return 0;
}