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(10 ** 7)
from bisect import bisect_left
def resolve():
    def dfs(s, parent):
        a = A[s]
        i = bisect_left(dp, a)
        v = dp[i]
        dp[i] = a
        ans[s] = bisect_left(dp, INF)
        for to in G[s]:
            if to == parent:
                continue
            dfs(to, s)
        # 頂点sでの結果をもとに戻す
        dp[i] = v
        return


    N = int(input())
    A = tuple(map(int, input().split()))
    G = [[] for _ in range(N)]
    for _ in range(N - 1):
        a, b = map(lambda x:int(x)-1, input().split())
        G[a].append(b)
        G[b].append(a)

    INF = 10 ** 18
    dp = [INF] * N
    ans = [0] * N

    dfs(0, -1)
    print(*ans, sep="\n")

if __name__ == "__main__":
    resolve()
