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
# -*- coding: utf-8 -*-
import sys
from bisect import bisect_left, bisect_right
sys.setrecursionlimit(10**6)

N=int(sys.stdin.readline().strip())
A=[None]+map(int, sys.stdin.readline().split())

al=[ [] for _ in range(N+1) ]	#隣接リスト
for i in range(N-1):
	u,v=map(int, sys.stdin.readline().split())
	al[u].append(v)
	al[v].append(u)

LIS=[ float("inf") for _ in range(N+1)] 
LIS[0]=float("-inf")
ans={}  #頂点i:LISの長さ

def dfs(cur,par):
    idx=bisect_left(LIS, A[cur])
    old=LIS[idx]    #LISの更新前の値をoldに持っておく
    LIS[idx]=A[cur]
    ans[cur]=bisect_left(LIS, float("inf"))-1 #LISの長さ
    for to in al[cur]:
        if par==to: continue    #遷移先が親であれば遷移しない
        dfs(to,cur)
    LIS[idx]=old    #oldの値を戻して、LIS配列を頂点curまで遷移した時点の状態に復元する
        
dfs(1,-1)

for i in range(1,N+1):
    print ans[i]