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.

-----Sample 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

-----Sample Output-----
1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import bisect\nimport sys\nsys.setrecursionlimit(10**7)\n\ndef dfs(v):\n #DFSの行き(子ノードに下っていくとき)の処理\n pos=bisect.bisect_left(dp,arr[v]) #最長増加部分列で更新する場所を2分探索により求める\n changes.append((pos,dp[pos])) #更新した要素とその値を記録しておく\n dp[pos]=arr[v]\n ans[v]=bisect.bisect_left(dp,10**18) #1からvまでの最長増加部分列の長さは、最長増加部分列の10**18以外の値の個数に等しい\n for u in g[v]:\n   if checked[u]==0:\n     checked[u]=1\n     dfs(u)\n #DFSの戻り(親ノードに上っていくとき)の処理\n pos,val=changes.pop() #頂点vで更新した最長増加部分列の値を元に戻す\n dp[pos]=val\n\nn=int(input())\narr=[0]+list(map(int,input().split()))\ng=[[] for _ in range(n+1)]\nfor _ in range(n-1):\n a,b=map(int,input().split())\n g[a].append(b)\n g[b].append(a)\nans=[0]*(n+1)\nchecked=[0]*(n+1)\nchecked[1]=1\ndp=[10**18 for _ in range(n+1)] #最長増加部分列を求めるのに、十分大きな値で初期化しておく\nchanges=[]\ndfs(1)\nfor i in range(1,n+1):\n print(ans[i])'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13547