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
input = lambda: sys.stdin.readline().rstrip()

n = int(input())
a = [int(x) for x in input().split()]
g = [[] for _ in range(n)]
for _ in range(n - 1):
    u, v = map(int, input().split())
    g[u - 1].append(v - 1)
    g[v - 1].append(u - 1)
from collections import defaultdict
import bisect
def dfs(u):
    ans = [-1]*n # 各最長増加部分列の長さ
    stack = [u] # u: 頂点
    operation = defaultdict(list) # 行った操作
    used = [False]*n # 1回目: False 2回目: True
    dp = [float("inf")]*n
    dp[0] = a[u]
    used[u] = False
    length = 1
    operation[0] = [0, float("inf"), 0]
    ans[0] = 1
    while stack:
        # print(dp)
        u = stack[-1]
        if used[u] == False:
            used[u] = True
            idx = bisect.bisect_left(dp, a[u])
            operation[u] = [idx, dp[idx], length]
            dp[idx] = a[u]
            length = max(length, idx + 1)
            ans[u] = length
            for i in g[u]:
                if ans[i] > 0:
                    continue
                stack.append(i)
        else:
            idx, value, length = operation[u]
            dp[idx] = value
            stack.pop()
    return ans, used, dp
ans, used, dp = dfs(0)
print(*ans, sep="\n")
# print(used, dp)
                
            