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
#!/usr/bin/env python3
#F

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))

n = I()
a = LI()
edges = [[] for _ in range(n)]
for _ in range(n-1):
    u, v = LI()
    edges[u-1].append(v-1)
    edges[v-1].append(u-1)

dp = [inf]*n
ans = [None]*n
ans[0] = 1
dp[0] = a[0]
def dfs(s, p, dp, num):
    for u in edges[s]:
        flag = False
        if u == p:
            continue
        r = bl(dp, a[u])
        tmp = dp[r]
        if dp[r] == inf:
            num += 1
            flag = True
        ans[u] = num
        dp[r] = min(dp[r], a[u])
        dfs(u, s, dp, num)
        dp[r] = tmp
        if flag:
            num -= 1
    return
dfs(0, -1, dp, 1)

for i in ans:
    print(i)
        