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
    import sys
    sys.setrecursionlimit(10 ** 7)
    input = sys.stdin.readline

    inf = 1 << 30

    N = int(input())
    *a, = map(int, input().split())

    g = tuple(set() for _ in range(N))
    for _ in range(N - 1):
        u, v = (int(x) - 1 for x in input().split())
        g[u].add(v)
        g[v].add(u)

    ans = [0] * N
    lis = [inf] * N

    def dfs(v, p=-1):
        i = bisect_left(lis, a[v])
        prev_at_i = lis[i]
        lis[i] = a[v]
        ans[v] = bisect_left(lis, inf)
        for u in g[v]:
            if p == u: continue
            dfs(u, v)
        lis[i] = prev_at_i

    dfs(0)

    print(*ans, sep='\n')


if __name__ == '__main__':
    main()

# dp[i]=長さiの増加部分列の末尾としてありえる最小値
# というテーブルを持っておいて、数列を列を前から見ながら二分探索で更新箇所を探す
