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
input = sys.stdin.buffer.readline

sys.setrecursionlimit(10 ** 7)

from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict
from itertools import combinations, permutations
from itertools import accumulate
from math import ceil, sqrt, pi

def LLL(a, ls):
    INF = 10 ** 15
    
    flag = 0
    prev = 0
    indx = 0
    ls_i = bisect_left(ls, a)
    if a < ls[ls_i]:
        flag = 1
        prev = ls[ls_i]
        indx = ls_i
    ls[ls_i] = min(ls[ls_i], a)

    ls_ii = bisect_left(ls, INF)
    return ls_ii, ls, flag, prev, indx

MOD = 10 ** 9 + 7
INF = 10 ** 18

N = int(input())
A = [0] + list(map(int, input().split()))
UV = [list(map(int, input().split())) for _ in range(N - 1)]

graph = [[] for _ in range(N + 1)]
for uv in UV:
    u, v = uv
    graph[u].append(v)
    graph[v].append(u)
#print(graph)

INF = 10 ** 15

LIS = [INF for i in range(N + 2)]

answer = [0] * (N + 1)
def dfs(prev, n, LIS):
    answer[n], LIS, flag, prv, indx = LLL(A[n], LIS)
    #print(n, LIS, flag, prv, indx)
    for g in graph[n]:
        if g != prev:
            dfs(n, g, LIS)
    if flag:
        LIS[indx] = prv

dfs(-1, 1, LIS)

for a in answer[1:]:
    print(a)