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.readline
import bisect

sys.setrecursionlimit(10**7)
n = int(input())
l = list(map(int,input().split()))
e = [[] for i in range(n)]
for i in range(n-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    e[a].append(b)
    e[b].append(a)


ans = [0]*n
ans[0] = 1
dp = [l[0]]

def dfs(now,bef):
    
    change = True
    if l[now] > dp[-1]:
        dp.append(l[now])
    else:
        index = bisect.bisect_left(dp,l[now])
        change = False
        s = dp[index]
        dp[index] = l[now] 
    ans[now] = len(dp)
    
    for nex in e[now]:
        if nex != bef:
            dfs(nex,now)
    if change:
        dp.pop()
    else:
        dp[index] = s 

for i in e[0]:
    dfs(i,0)

for i in ans:
    print(i)

