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
sys.setrecursionlimit(10**6)
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
from bisect import bisect_left
def resolve():
    N = ir()
    a = lr()
    tr = [[] for _ in range(N)]
    for i in range(N-1):
        u, v = lr()
        tr[u-1].append(v-1)
        tr[v-1].append(u-1)
    INF = pow(10, 9)+1
    lis = [INF]*N
    lis[0] = a[0]
    ans = [0]*N
    ans[0] = 1
    def dfs(v, p=-1):
        for u in tr[v]:
            if u == p:
                continue
            idx = bisect_left(lis, a[u])
            prev = lis[idx]
            lis[idx] = a[u]
            ans[u] = bisect_left(lis, INF)
            dfs(u, v)
            lis[idx] = prev
    dfs(0)
    print(*ans, sep='\n')
resolve()