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
# coding: utf-8
# Your code here!
import sys
readline = sys.stdin.readline
read = sys.stdin.read

n = int(input())
a = [int(i) for i in readline().split()]
m = map(int,read().split())

g = [[] for _ in range(n)]
for i,j in zip(m,m):
    g[i-1].append(j-1)
    g[j-1].append(i-1)

INF = 10**9

q = [~0,0]
dp = [INF]*n
ans = [0]*n
val = 0
idx = [-1]*n
old = [-1]*n
parent = [-1]*n

ss = 0
from bisect import bisect_left
while q:
    v = q.pop()
    if v >= 0:
        iv = idx[v] = bisect_left(dp,a[v])
        x = old[v] = dp[idx[v]]
        if x == INF: val += 1
        ans[v] = val

        dp[iv] = a[v]
        for c in g[v]:
            if c != parent[v]:
                parent[c] = v            
                q.append(~c)
                q.append(c)
    else:
        v = ~v
        x = dp[idx[v]] = old[v]
        if x == INF: val -= 1
    

print(*ans,sep="\n")

