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
from collections import deque
import sys
import bisect
def input(): return sys.stdin.readline().rstrip()

n=int(input())
A=[0]+list(map(int, input().split()))

graph = [[] for _ in range(n+1)]

for i in range(n-1):
    a, b = map(int, input().split())
    graph[a].append(b)
    graph[b].append(a)

visited = [-1] * (n+1)
visited[0] = 0
visited[1] = 1

d = deque()
d.append(1)
dp=[10**10]*(n+1)
dp[0]=0
dp[1]=A[1]
update_task=[() for _ in range(n+1)]
update_task[1]=(1,0)
while d:
    v = d[-1]
    if graph[v] == []:
        d.pop()
        index,atai=update_task[v]
        dp[index]=atai
    else:
        i = graph[v].pop()
        if visited[i] != -1:
            continue
        bis = bisect.bisect_left(dp, A[i])
        update_task[i]=(bis,dp[bis])
        dp[bis] = A[i]
        visited[i] = bisect.bisect_left(dp, 10**10-100)-1
        d.append(i)

print(*visited[1:], sep="\n")