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
import bisect  # 二分法


MAX_N = 200005
INF = 10**9 + 5

sys.setrecursionlimit(MAX_N)

N = int(sys.stdin.readline())
A = [int(x) for x in sys.stdin.readline().split()]
E = [[] for _ in range(N)]  # 辺の情報

for _ in range(N-1):
    u, v = map(int, sys.stdin.readline().split())
    E[u-1].append(v-1)
    E[v-1].append(u-1)

lis = [INF] * N
ans = [0] * N 

def dfs(u,r=-1):

    # u を訪れたときの処理
    i = bisect.bisect_left(lis, A[u])  # 
    old = lis[i]  # 書き換え前の値を一時保存
    lis[i] = A[u]

    for v in E[u]:
        if v == r: continue  # 親は無視

        dfs(v,u)
    
    ans[u] = bisect.bisect_left(lis,INF-1)
    lis[i] = old  # 書き換え前の状態に戻す

dfs(0)
for i in range(N):
    print(ans[i])