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
sys.setrecursionlimit(10**7)
from bisect import bisect_left
def main(N,a,tr):
  ans=[0]*N
  dp=[float('inf')]*(N+1)
  dp[0]=0
  maxa=max(a)+1
  def dfs(p,v,dp):
    i=bisect_left(dp,a[v])
    tmp=dp[i]
    dp[i]=a[v]
    ans[v]=bisect_left(dp,maxa)-1
    for nv in tr[v]:
      if nv!=p:
        dfs(v,nv,dp)
    dp[i]=tmp
  dfs(-1,0,dp)
  print(*ans,sep='\n')
if __name__=='__main__':
  N=int(input())
  a=list(map(int,input().split()))
  tr=[[] for _ in range(N)]
  for _ in range(N-1):
    u,v=map(int,input().split())
    u,v=u-1,v-1
    tr[u].append(v)
    tr[v].append(u)
  main(N,a,tr)
