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
import math
import heapq
import bisect
sys.setrecursionlimit(10**7)
INTMAX = 9223372036854775807
INTMIN = -9223372036854775808
DVSR = 1000000007
def POW(x, y): return pow(x, y, DVSR)
def INV(x, m=DVSR): return pow(x, m - 2, m)
def DIV(x, y, m=DVSR): return (x * INV(y, m)) % m
def LI(): return map(int, sys.stdin.readline().split())
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())


def FLIST(n):
    res = [1]
    for i in range(1, n+1):
        res.append(res[i-1]*i % DVSR)
    return res


N = II()
AS = list(LI())

MP = {}

MX = 10**10
LIS = [10**10]*N
RES = [1]*N
for i in range(N-1):
    a, b = LI()
    a -= 1
    b -= 1
    if not a in MP:
        MP[a] = []
    if not b in MP:
        MP[b] = []
    MP[a].append(b)
    MP[b].append(a)


def dfs(v, mp, lis, p):
    for u in mp[v]:
        if u != p:
            i = bisect.bisect_left(lis, AS[u])
            bef = lis[i]
            lis[i] = AS[u]
            longest = bisect.bisect_left(lis, MX)
            RES[u] = longest
            dfs(u, mp, lis, v)
            lis[i] = bef


LIS[0] = AS[0]
dfs(0, MP, LIS, -1)

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