human
Write me a Python program to find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators given the following:
I found an interesting problem on https://codeforces.com/problemset/problem/935/E:
Ancient Egyptians are known to have understood difficult concepts in mathematics. The ancient Egyptian mathematician Ahmes liked to write a kind of arithmetic expressions on papyrus paper which he called as Ahmes arithmetic expression.

An Ahmes arithmetic expression can be defined as:   "d" is an Ahmes arithmetic expression, where d is a one-digit positive integer.
 "(E_1 op E_2)" is an Ahmes arithmetic expression, where E_1 and E_2 are valid Ahmes arithmetic expressions (without spaces) and op is either plus ( + ) or minus ( - ).  For example 5, (1-1) and ((1+(2-3))-5) are valid Ahmes arithmetic expressions.

On his trip to Egypt, Fafa found a piece of papyrus paper having one of these Ahmes arithmetic expressions written on it. Being very ancient, the papyrus piece was very worn out. As a result, all the operators were erased, keeping only the numbers and the brackets. Since Fafa loves mathematics, he decided to challenge himself with the following task:

Given the number of plus and minus operators in the original expression, find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators.

-----Input-----

The first line contains a string E (1 ≤ |E| ≤ 10^4) — a valid Ahmes arithmetic expression. All operators are erased and replaced with '?'.

The second line contains two space-separated integers P and M (0 ≤ min(P, M) ≤ 100) — the number of plus and minus operators, respectively. 

It is guaranteed that P + M =  the number of erased operators.

-----Output-----

Print one line containing the answer to the problem.

-----Examples-----
Input
(1?1)
1 0

Output
2

Input
(2?(1?2))
1 1

Output
1

Input
((1?(5?7))?((6?2)?7))
3 2

Output
18

Input
((1?(5?7))?((6?2)?7))
2 3

Output
16


-----Note-----  The first sample will be (1 + 1)  =  2.  The second sample will be (2 + (1 - 2))  =  1.  The third sample will be ((1 - (5 - 7)) + ((6 + 2) + 7))  =  18.  The fourth sample will be ((1 + (5 + 7)) - ((6 - 2) -...

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python program to find out the maximum possible value for the expression on the papyrus paper after putting the plus and minus operators in the place of the original erased operators.

from math import inf

class Node:
    def __init__(self, parent = None, leftExp = None, rightExp = None, signQ = 0):
        self.parent, self.leftExp, self.rightExp, self.signQ = parent, leftExp, rightExp, signQ
    def __str__(self):
        return "Node"

memo = {}

def Memoize(node, p, maxValue, minValue):
    if not node in memo:
        memo.update({node : {p : [maxValue, minValue]} })
    else:
        memo[node].update({p : [maxValue, minValue]})

def ExpMaxValue(root: Node, p):

    m = root.signQ - p
    """if root.signQ == 1:
        if p == 1:
            return [root.leftExp.value + root.rightExp.value, root.leftExp.value + root.rightExp.value]
        else:
            return [root.leftExp.value - root.rightExp.value, root.leftExp.value - root.rightExp.value]"""

    if root.signQ == 0:
        return [root.value, root.value]
    
    if root in memo:
        if p in memo[root]:
            return memo[root][p]

    if m == 0:
        value = ExpMaxValue(root.leftExp, root.leftExp.signQ)[0] + ExpMaxValue(root.rightExp, root.rightExp.signQ)[0]
        Memoize(root, p, value, value)
        return [value, value]
    if p == 0:
        value = ExpMaxValue(root.leftExp, 0)[0] - ExpMaxValue(root.rightExp, 0)[0]
        Memoize(root, p, value, value)
        return [value, value]
        
    maxValue = -inf
    minValue = inf
    if m >= p:
        for pQMid in range(2):
            pQLeftMin = min(p - pQMid, root.leftExp.signQ)
            for pQLeft in range(pQLeftMin + 1):
                if root.leftExp.signQ - pQLeft + (1 - pQMid) > m:
                    continue
                
                resLeft = ExpMaxValue(root.leftExp, pQLeft)
                resRight = ExpMaxValue(root.rightExp, p - pQMid - pQLeft)
                
                if pQMid == 1:
                    maxValue = max(resLeft[0] + resRight[0], maxValue)
                    minValue = min(resLeft[1] + resRight[1], minValue)
                else:
                    maxValue = max(resLeft[0] - resRight[1], maxValue)
     ...