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
'''
自宅用PCでの解答
'''
import math
#import numpy as np
import itertools
import queue
import bisect
from collections import deque,defaultdict
import heapq as hpq
from sys import stdin,setrecursionlimit
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse import csr_matrix
ipt = stdin.readline
setrecursionlimit(10**7)



def main():
    n = int(ipt())
#    n = 2*10**5
    a = [int(i) for i in ipt().split()]
#    a = [i+1 for i in range(n)]
    way = [[] for i in range(n)]
    for i in range(n-1):
        u,v = map(int,ipt().split())
        u -= 1;v -= 1
        way[u].append(v)
        way[v].append(u)
#    way = [[i-1,i+1] for i in range(n)]
#    way[0] = [1]
#    way[n-1] = [n-2]
    ans = [0]*n
    dp = [10**18]*n
    stk = []

    def solve(ni,pi):
        ai = a[ni]
        pt = bisect.bisect_left(dp,ai)
        stk.append((pt,dp[pt]))
        dp[pt] = ai
        ans[ni] = bisect.bisect_left(dp,10**18-1)
        for i in way[ni]:
            if i == pi:
                continue
            else:
                solve(i,ni)
        pl,val = stk.pop()
        dp[pl] = val
        return None

    solve(0,-1)

    for i in ans:
        print(i)

    return None

if __name__ == '__main__':
    main()
