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 copy import copy
from bisect import bisect_left, bisect_right
input = sys.stdin.readline
sys.setrecursionlimit(10**9)

n = int(input())
a = list(map(int, input().split()))

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

ans = [0] * n
dp = [float('inf')] * n
dp[0] = a[0]


def dfs(v, v_p, dp_len):
    # print(v, v_p, dp)
    ans[v] = dp_len
    for v_next in graph[v]:
        if v_next == v_p:
            continue

        # print(dp)
        if a[v_next] > dp[dp_len - 1]:
            past = dp[dp_len]
            dp[dp_len] = a[v_next]
            dfs(v_next, v, dp_len + 1)
            dp[dp_len] = past
        else:
            idx = bisect_left(dp, a[v_next])
            # always idx < dp_len
            past = dp[idx]
            dp[idx] = a[v_next]
            dfs(v_next, v, dp_len)
            dp[idx] = past


dfs(0, -1, 1)
print(*ans, sep='\n')
