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**7)

import sys
input = sys.stdin.readline

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

from collections import defaultdict
G = defaultdict(list)

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

from bisect import bisect
ansl = [0 for i in range(n)]
dp = [float("inf") for _ in range(n+1)]
dp[0] = -1
used = [0 for i in range(n)]
used[0] = 1


stack = []
def dfs(pos):
    idx = bisect(dp, al[pos] - 1)
    stack.append((idx, dp[idx]))
    dp[idx] = min(al[pos], dp[idx])
    ansl[pos] = bisect(dp, 10**18) - 1
    for v in G[pos]:
        if used[v] == 0:
            used[v] = 1
            dfs(v)
            idx, k = stack.pop()
            dp[idx] = k


dfs(0)
flag = False
for i in range(n):
    print(ansl[i])


