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
sys.setrecursionlimit(10**8)
N = int(input())
A = list(map(int,input().split()))
UV = [tuple(map(int,input().split())) for i in range(N-1)]
es = [[] for _ in range(N)]
for u,v in UV:
    u,v = u-1,v-1
    es[u].append(v)
    es[v].append(u)

from bisect import bisect_left
lis = [A[0]]
stack = []
ans = [1] * N
def dfs(v,p=-1):
    for to in es[v]:
        if to==p: continue
        i = bisect_left(lis, A[to])
        if i == len(lis):
            stack.append((i,-1))
            lis.append(A[to])
        else:
            stack.append((i,lis[i]))
            lis[i] = A[to]
        ans[to] = len(lis)

        dfs(to,v)

        i,a = stack.pop()
        if a < 0:
            lis.pop()
        else:
            lis[i] = a

dfs(0)
print(*ans, sep='\n')