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 bisect
import sys
sys.setrecursionlimit(10**9)

N = int(input())
A = list(map(int, input().split()))

tree = [[] for i in range(N)]
for i in range(N-1):
  u, v = map(int, input().split())
  tree[u-1].append(v-1); tree[v-1].append(u-1)


ans = [0] * N
ans[0] = 1
INF = 10**10
dp = [INF] * N
dp[0] = A[0]
stack = []
n_stack = []
def dfs(node, par, n):
  global ans
  global dp
  global stack
  children = tree[node]
  for child in children:
    if child == par:
      continue
    num = A[child]
    idx = bisect.bisect_left(dp, num)
    stack.append([idx, dp[idx]])
    n_stack.append(n)
    dp[idx] = num
    n = max(n, idx + 1)
    ans[child] = n
    dfs(child, node, n)
    # 巻き戻し
    idx, num = stack.pop()
    dp[idx] = num
    n = n_stack.pop()

dfs(0, -1, 1)
print(*ans, sep='\n')