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
# F - LIS on Tree

from bisect import bisect_left

n = int(input())
a = list(map(int, input().split()))
assert len(a) == n
adj = {u: [] for u in range(n)}
for i in range(n - 1):
    u, v = map(int, input().split())
    adj[u - 1].append(v - 1)
    adj[v - 1].append(u - 1)

# lis[i] == 長さi+1の増加列の末尾の最小値
lis = [float('inf')] * n

lislen = [0] * n  # lislen[u] == 頂点uまでのLISの長さ
backup = [None] * n

# DFS (with an explicit stack)
stack = [(0, 0)]  # (parent, child)
while len(stack) > 0:
    p, u = stack.pop()
    if p == None:  # 帰りがけ
        l, x = backup[u]
        lis[l] = x  # 復元
    else:  # 行きがけ
        l = bisect_left(lis, a[u])  # a[u]を使って長さl+1のLISを作れる
        backup[u] = (l, lis[l])  # 後で復元するために保存
        lis[l] = a[u]  # 更新
        lislen[u] = bisect_left(lis, float('inf'))  # 現時点のLISの長さ
        stack.append((None, u))  # 帰りがけタスクをpush
        for v in adj[u]:
            if v != p:
                stack.append((u, v))

for u in range(n):
    print(lislen[u])
