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(100000000)
from  bisect import bisect_left
input = sys.stdin.readline
INF = 1 << 30

MAXN = 200005
G = [[] for _ in range(MAXN)]
dp = [INF] * MAXN
stack = []
ans = [None] * MAXN
def dfs(v,A,p = -1):
    idx = bisect_left(dp,A[v])
    stack.append((idx,dp[idx]))
    dp[idx] = A[v]
    ans[v] = bisect_left(dp,INF)
    for e in G[v]:
        if e == p:
            continue
        dfs(e,A,v)
    idx,p = stack.pop()
    dp[idx] = p

def main():
    N = int(input())
    A = list(map(int,input().split()))
    for _ in range(N - 1):
        a,b = map(int,input().split())
        a -= 1
        b -= 1
        G[a].append(b)
        G[b].append(a)
    dfs(0,A)
    print('\n'.join(map(str,ans[:N])))
if __name__ == '__main__':
    main()
    

    
    
