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 defaultdict as dd
N = int(input())
As = list(map(int, input().split()))
Es = dd(dict)
for _ in range(N-1):
    f, t = map(int, input().split())
    Es[f-1][t-1] = Es[t-1][f-1] = 1

INF = float('inf')
RET = 0
PROC = 1

stack = []
lismin = [INF] * N
anss = [INF] * N
visited = [False] * N

stack.append((RET, 0, INF))
stack.append((PROC, 0, 0))
while stack:
    cmd, *v = stack.pop()
    if cmd == RET:
        i, backup = v
        lismin[i] = backup
    else:
        node, i = v
        lismin[i] = As[node]
        anss[node] = bisect_left(lismin, INF)
        visited[node] = True
        for to in Es[node].keys():
            if not visited[to]:
                x = bisect_left(lismin, As[to])
                stack.append((RET, x, lismin[x]))
                stack.append((PROC, to, x))
for ans in anss:
    print(ans)

