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 = '# coding: utf-8\n# Your code here!\nimport sys\nreadline = sys.stdin.readline\nread = sys.stdin.read\n\nN = int(input())\nA = list(map(int, input().split()))\nG = [[] for _ in range(N)]\n\nfor i in range(N-1):\n  a,b = list(map(int, input().split()))\n  G[a-1].append(b-1)\n  G[b-1].append(a-1)\n\nINF = 10**9+1\n\n#a = [~0,0]\nq = [0]    # ~0=-1なのだが、とにかく0を打ち消す方向をやるよ、と言うこと。うまい。\ndp = [INF]*N  # LIS\nans = [0]*N   # iにおけるLISの長さ\nlis_len = 0   # 持ち回ってるLIS長さ変数。dpをsearchしても求まるけど、安易に野郎\nidx = [-1]*N  # iのときの操作したIDX\nold = [-1]*N  # iのときに操作する前の値。Rollbackする時に使う\nprev = [-1]*N # 自分の呼び出しもと。グラフの行ってこいを抑止するため。\n\nss = 0\nfrom bisect import bisect_left\nwhile q:\n  now = q.pop()\n  # 普通のdfs\n  if now >= 0:\n    a = A[now]\n    # LIS管理のdpに今回操作する対象のIDXを探索\n    idx[now] = bisect_left(dp,a)\n    # 操作前の値とIDXを保存する\n    iv = idx[now]\n    old[now] = dp[idx[now]]\n    x = old[now]\n    # LIS管理dpのVALUEがINFだったら、今回で長さが伸びるってこと\n    if x == INF:\n      lis_len += 1\n    dp[iv] = a\n    ans[now] = lis_len\n\n    for nxt in G[now]:\n      if nxt == prev[now]:\n        continue\n      prev[nxt] = now\n      #q.append(~nxt)\n      #q.append(nxt)\n      q.append((-1)*nxt)\n      q.append(nxt)\n  # nowがゼロ以下なので、DPを一つRollbackする動きをする。\n  else:\n    #v = ~v # ビット反転だが、これで元のNOWを復元する\n    now = (-1)*now\n    dp[idx[now]] = old[now]\n    x = dp[idx[now]] \n    if x == INF:\n      lis_len -= 1\n    \n#print(*ans,sep="\\n")\nfor a in ans:\n  print(a)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13959