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 bisect
import sys
sys.setrecursionlimit(500000)

n = int(input())
a = list(map(int, input().split()))
to=[list() for _ in range(n)]

for i in range(n-1):
    u,v = map(int, input().split())
    u-=1
    v-=1
    to[u].append(v)
    to[v].append(u)

INF=1001001001
dp = [INF]*n
dp[0]=a[0]
ansls=[0]*n
ansls[0]=1
history=deque()
def dfs(u,p,rank):
    for v in to[u]:
        if v==p:
            continue
        idx=bisect.bisect_left(dp,a[v])
        prv=dp[idx]
        dp[idx]=a[v]       
        history.append((idx,prv))
        ansls[v]=bisect.bisect_left(dp,INF)
        dfs(v,u,rank+1)
        while len(history)>rank:
            idx,prv =history.pop()
            dp[idx] = prv

dfs(0,-1,0)

print(*ansls,sep='\n')