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)


def input():
    return sys.stdin.readline()


n = int(input())
aa = [int(i) for i in input().split()]
e = [[] for _ in range(n)]
for _ in range(n - 1):
    a, b = [int(i) - 1 for i in input().split()]
    e[a].append(b)
    e[b].append(a)

dp = [10 ** 9] * (n + 10)
dp[0] = 0
ans1 = [0] * n
ans2 = [0] * n


def dfs(i=0, r=-1):
    global dp, ans1, ans2
    mi = 0
    ma = n
    while mi != ma:
        mm = (mi + ma) // 2
        if dp[mm] >= aa[i]:
            ma = mm
        else:
            mi = mm + 1

    be = dp[mi]
    dp[mi] = min(dp[mi], aa[i])
    ans1[i] = mi
    ans2[i] = max(ans1[i], ans2[r])

    for j in e[i]:
        if j == r:
            continue
        dfs(j, i)

    dp[mi] = be


dfs()
print("\n".join(map(str, ans2)))
