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(2147483647)
INF=float("inf")
MOD=10**9+7 # 998244353
input=lambda:sys.stdin.readline().rstrip()
from bisect import bisect_left
def resolve():
    n = int(input())
    A = list(map(int,input().split()))
    E = [[] for _ in range(n)]
    for _ in range(n - 1):
        u, v = map(int,input().split())
        u -= 1; v -= 1
        E[u].append(v)
        E[v].append(u)

    dp = [INF] * (n + 1)
    dp[0] = -INF
    res = 0
    ans = [None] * n
    change = [None] * n
    stack = [(0, -1)]

    while stack:
        v, p = stack.pop()
        if v >= 0:
            idx = bisect_left(dp, A[v])
            change[v] = (idx, dp[idx])
            res = max(res, idx)
            ans[v] = res
            dp[idx] = A[v]
            for nv in E[v]:
                if nv == p:
                    continue
                stack.append((~nv, v))
                stack.append((nv, v))
        else:
            v = ~v
            idx, prev = change[v]
            dp[idx] = prev
            res = ans[p]

    print(*ans)
resolve()