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**6)

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

E=[[] for i in range(N+1)]
for i in range(N-1):
    x,y=map(int,input().split())
    E[x].append(y)
    E[y].append(x)

USE=[0]*(N+1)
USE[1]=1
ANS=[0]*(N+1)
DP=[]

def dfs(x):
    pos=bisect.bisect_left(DP,A[x])
    
    if pos<len(DP):
        changepos=pos
        old=DP[pos]
        
        DP[pos]=A[x]
    else:
        changepos=pos
        old=-1
        
        DP.append(A[x])

    ANS[x]=len(DP)

    for to in E[x]:
        if USE[to]==0:
            USE[to]=1
            dfs(to)

    if old==-1:
        DP.pop()
    else:
        DP[changepos]=old

dfs(1)

print(*ANS[1:])