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 bisect import bisect_left
from collections import deque

N = int(input())
A = list(map(int, input().split()))
X = [list(map(int, input().split())) for _ in range(N - 1)]

graph = [[] for _ in range(N)]
for u, v in X:
    graph[u - 1].append(v - 1)
    graph[v - 1].append(u - 1)

INF = 10 ** 9 + 7
dp = [INF] * (N + 1)
dp[0] = -INF
rb = [[] for _ in range(N)]
ans = [0] * N

# DFS
q = deque([(0, 0)])  # (current, parent)
while q:
    u, p = q.pop()

    # Roll back
    if p is None:
        i, x = rb[u].pop()
        dp[i] = x
        continue

    # Update dp
    i = bisect_left(dp, A[u])
    rb[u].append((i, dp[i]))
    dp[i] = A[u]

    # Calc ans
    ans[u] = max(i, ans[p])

    # Go to next
    q.append((u, None))
    for v in graph[u]:
        if v == p:
            continue
        q.append((v, u))

print(*ans)
