Beat 100% || Very very easy solution || O(n*m)
Code

# Code
```python3 []
class Solution:
    def maximumAmount(self, coins: List[List[int]]) -> int:
        n, m = len(coins), len(coins[0])

        dp = []
        for i in range(n+1):
            raw_lst = []
            for j in range(m+1):
                lst = []
                for k in range(3):
                    lst.append(float('-inf') if i == n or j == m else 0)
                raw_lst.append(lst)
            dp.append(raw_lst)
        dp[-1][-2] = [0, 0, 0]
        dp[-2][-1] = [0, 0, 0]
                
                    
     
                
        for i in range(n - 1, -1, -1):
            for j in range(m - 1, -1, -1):
                for k in range(2, 0, -1):
                    if coins[i][j] >= 0:
                        dp[i][j][k] = coins[i][j] + max(dp[i+1][j][k], dp[i][j+1][k])
                    else:
                        take = max(dp[i+1][j][k-1], dp[i][j+1][k-1])
                        not_take = coins[i][j] + max(dp[i+1][j][k], dp[i][j+1][k])
                        dp[i][j][k] = max(take, not_take)
                dp[i][j][0] = coins[i][j] + max(dp[i+1][j][0], dp[i][j+1][0])

        return dp[0][0][2]
                    


            
```
['Python3']