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
import sys
sys.setrecursionlimit(1000000)
from bisect import bisect, bisect_left
def dfs(vertex):
  visited[vertex] = True
  value = a[vertex]
  j = bisect(dp, value)
  previous = dp[j]
  if dp[j-1] != value: # strict increasing
    dp[j] = value
  ans[vertex] = bisect_left(dp, float("inf"))-1
  for node in adjacent[vertex]:
    if not visited[node]:
      dfs(node)
  dp[j] = previous
  return
n = int(input())
a = list(map(int, input().split()))
adjacent = {i: [] for i in range(n)}
for _ in range(n-1):
  node1, node2 = map(int, input().split())
  node1 -= 1
  node2 -= 1
  adjacent[node1].append(node2)
  adjacent[node2].append(node1)
visited = [False]*n
ans = [0]*n
dp = [float("inf") for _ in range(n+1)]
dp[0] = float("-inf")
dfs(0)
[print(value) for value in ans]