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 <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

const int INF = 0x3f3f3f3f;

const int N = 200000;
int n;
std::vector<int> g[N];
int a[N], ans[N];

int l[N];
void dfs(int v, int pr = -1) 
{
	static int Ans = 0;
	int *p = std::lower_bound(l, l + n, a[v]);
	int prev = *p;
	if (prev == INF) ++Ans;
	*p = a[v];
	ans[v] = Ans;
	for (int i = 0; i < (int)g[v].size(); ++i) if (g[v][i] != pr) dfs(g[v][i], v);
	if (prev == INF) --Ans;
	*p = prev;
}

int main() 
{
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) scanf("%d", a + i);
	for (int i = 0; i < n - 1; ++i) {
		int a, b;
		scanf("%d%d", &a, &b);
		g[a - 1].push_back(b - 1);
		g[b - 1].push_back(a - 1);
	}
	memset(l, 0x3f, 4 * n);
	dfs(0);
	for (int i = 0; i < n; ++i) printf("%d\n", ans[i]);
	return 0;
}
