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 sys
sys.setrecursionlimit(1000000)

input = sys.stdin.readline

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

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

ans = [0] * n
visited = [0] * n
visited[0] = 1
l = []

def dfs(i):
    s = a[i]
    pi = -1
    pv = -1
    if len(l) == 0 or s > l[-1]:
        l.append(s)
    else:
        pi = bisect.bisect_left(l, s)
        pv = l[pi]
        l[pi] = s
    ans[i] = len(l)

    for ni in tree[i]:
        if not visited[ni]:
            visited[ni] = 1
            dfs(ni)
            
    if pi == -1:
        l.pop()
    else:
        l[pi] = pv


dfs(0)

for i in ans:
    print(i)
