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 f_lis_on_tree(INF=float('inf')):
    # 参考: https://note.com/tanon_cp/n/n51ee6db8f5b2
    import sys
    sys.setrecursionlimit(10**7)
    import bisect
    N = int(input())
    A = [0] + [int(i) for i in input().split()]  # 以下、1-origin で考える
    Edges = [[int(i) for i in input().split()] for j in range(N - 1)]

    graph = [[] for _ in range(N + 1)]
    for a, b in Edges:
        graph[a].append(b)
        graph[b].append(a)

    ans = [0] * (N + 1)
    visited = [False] * (N + 1)
    visited[1] = True  # 根は訪問済
    lis = [INF] * (N + 1)
    changes = []

    def dfs(v):
        # 行きがけの処理
        # LIS の更新する場所を二分探索で求める
        pos = bisect.bisect_left(lis, A[v])
        changes.append((pos, lis[pos]))  # 更新した要素とその値を記録しておく
        lis[pos] = A[v]
        ans[v] = bisect.bisect_left(lis, INF)  # INF が現れるまでの要素数が v での解

        # 次の頂点へ
        for child in graph[v]:
            if not visited[child]:
                visited[child] = True
                dfs(child)

        # 帰りがけの処理 (頂点 v で更新した LIS の値を復元)
        pos, val = changes.pop()
        lis[pos] = val

    dfs(1)
    return ' '.join(map(str, ans[1:]))

print(f_lis_on_tree())