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
from bisect import bisect_left

sys.setrecursionlimit(10 ** 6)


def dfs(v, p, links, state, ans):
    ans[v] = len(state)
    for u in links[v]:
        if u == p:
            continue
        a = aaa[u]
        b, i = -1, -1
        if state[-1] < a:
            state.append(a)
        else:
            i = bisect_left(state, a)
            b = state[i]
            state[i] = a
        dfs(u, v, links, state, ans)
        if b == -1:
            state.pop()
        else:
            state[i] = b


n = int(input())
aaa = list(map(int, input().split()))
links = [set() for _ in range(n)]
for line in sys.stdin:
    u, v = map(int, line.split())
    u -= 1
    v -= 1
    links[u].add(v)
    links[v].add(u)
ans = [0] * n
dfs(0, -1, links, [aaa[0]], ans)
print('\n'.join(map(str, ans)))
