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 = 'n = int(input())\na = list(map(int, input().split()))\n\nuv = [[] for _ in range(n)]\nfor _ in range(n-1):\n    i, j = map(int, input().split())\n    uv[i-1].append(j-1)\n    uv[j-1].append(i-1)\n\n# DFS\n# 頂点iが未探索の時、ans[i] = -1\nans = [-1 for _ in range(n)]\nans[0] = 1\n# dp[i] = 長さi+1のLISの最後尾の最小値\ndp = [a[0]]\n# history[(i, j)]のとき、変更前はdp[i] = j\n# stackとして使用\nhistory = []\n# 頂点parents[i] = iの親\nparents = [-1 for _ in range(n)]\n\ntodo = []\nfor i in uv[0]:\n    todo.append(i)\n    parents[i] = 0\n\nwhile True:\n    i = todo.pop(-1)\n    \n    # ans[i]を求める\n    if dp[-1] < a[i]:\n        dp.append(a[i])\n        history.append((-1, 0))\n    else:\n        # dpの変更点を二分探索\n        start = 0\n        stop = len(dp) - 1\n        flag = True\n        while start < stop:\n            center = (start + stop) // 2\n            if dp[center] < a[i]:\n                start = center + 1\n            elif dp[center] > a[i]:\n                stop = center\n            else:\n                flag = False\n                break\n        if flag and dp[start] > a[i]:\n            history.append((start, dp[start]))\n            dp[start] = a[i]\n        else:\n            history.append((-2, 0))\n    ans[i] = len(dp)\n\n    # todoに追加\n    for j in uv[i]:\n        if ans[j] == -1:\n            todo.append(j)\n            parents[j] = i\n\n    if len(todo) == 0:\n        break\n\n    # back\n    j = i\n    while parents[todo[-1]] != j:\n        k = history.pop(-1)\n        if k[0] == -2:\n            None\n        elif k[0] == -1:\n            dp.pop(-1)\n        else:\n            dp[k[0]] = k[1]\n        j = parents[j]\n\nfor i in ans:\n    print(i)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14317