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
def main():

    from bisect import bisect_left as bl

    n = int(input())
    a = list(map(int, input().split()))
    ab = [list(map(int, input().split())) for _ in [0]*(n-1)]
    g = [set() for _ in [0]*n]
    ans = [0]*n
    [g[a-1].add(b-1) for a, b in ab]
    [g[b-1].add(a-1) for a, b in ab]
    visited = [False]*n

    q = [0]
    lis = []
    while q:
        i = q.pop()
        ai = a[i]
        if not visited[i]:
            b = bl(lis, ai)
            if b == len(lis):
                visited[i] = "push"
                lis.append(ai)
            else:
                visited[i] = [b, lis[b]]
                lis[b] = ai
            ans[i] = len(lis)
        if not g[i]:
            if visited[i] == "push":
                lis.pop()
            else:
                lis[visited[i][0]] = visited[i][1]
        else:
            q.append(i)
            j = g[i].pop()
            g[j].remove(i)
            q.append(j)
    for i in ans:
        print(i)


main()