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
from collections import defaultdict
from bisect import bisect_left
import sys
sys.setrecursionlimit(10 ** 6)

N = int(input())
A = list(map(int, input().split()))
dp = [float('inf')] * N
tree = defaultdict(list)
for _ in range(N - 1):
	u, v = list(map(int, input().split()))
	tree[u-1].append(v-1)
	tree[v-1].append(u-1)

dp[0] = A[0]
ans = [float('inf')] * N
ans[0] = 1
def backtrack(node, visited, tail_idx):
	flag = any([nx not in visited for nx in tree[node]])
	if not flag:
		return
	for nx in tree[node]:
		if nx in visited:
			continue
		visited.add(nx)
		idx = bisect_left(dp, A[nx])
		prev = dp[idx]
		if tail_idx < idx:
			dp[idx] = A[nx]
			new_tail_idx = idx
			ans[nx] = min(new_tail_idx + 1, ans[nx])
			backtrack(nx, visited, new_tail_idx)
		elif A[nx] < dp[idx]:
			dp[idx] = A[nx]
			ans[nx] = min(tail_idx + 1, ans[nx])
			backtrack(nx, visited, tail_idx)
		else:
			ans[nx] = min(tail_idx + 1, ans[nx])
			backtrack(nx, visited, tail_idx)
		dp[idx] = prev
		visited.remove(nx)

backtrack(node=0, visited={0}, tail_idx=0)
for i in ans:
	print(i)