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 bisect
import os

import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353


N = int(sys.stdin.buffer.readline())
A = list(map(int, sys.stdin.buffer.readline().split()))
VU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N - 1)]

graph = [[] for _ in range(N)]
for v, u in VU:
    v -= 1
    u -= 1
    graph[v].append(u)
    graph[u].append(v)

dp = [INF] * (N + 10)
ans = [0] * N
root = 0


def dfs(v, p):
    i = bisect.bisect_left(dp, A[v])
    prev = dp[i]
    dp[i] = A[v]
    ans[v] = bisect.bisect_left(dp, INF)
    # print(dp)
    for u in graph[v]:
        if u == p:
            continue
        dfs(u, v)
    assert dp[i] <= prev
    dp[i] = prev
    # print(dp)


dfs(root, None)
print(*ans, sep='\n')
