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
#F
import bisect
import sys

sys.setrecursionlimit(202020)

inf = float("inf")
N = int(input())
A = list(map(int,input().split()))

on = [[] for _ in range(N)]
for _ in range(N-1):
    u,v = map(int,input().split())
    on[u-1].append(v-1)
    on[v-1].append(u-1)
    
    

DP = [inf]*(N+1)
stack = []
ans = [-1]*N
seen = [False]*N

#DFS
def dfs(gragh, v): #v = start
    """
    global A
    global DP
    global ans
    global stack 
    global seen
    """
    
    seen[v] = True
    ind = bisect.bisect_left(DP,A[v])
    mae = DP[ind]
    DP[ind] = A[v]
    stack.append([ind,mae])
    
    ans[v] = bisect.bisect_left(DP,inf)
    
    for next_v in gragh[v]:
        if seen[next_v]: 
            continue
        
        dfs(gragh,next_v)
        
    ind,mae = stack.pop()
    DP[ind] = mae
    
    return



dfs(on,0)
for a in ans:
    print(a)

