class Solution:
    def maximumAmount(self, coins: List[List[int]]) -> int:
        m, n = len(coins), len(coins[0])
        dp = [[-math.inf] * 3 for _ in range(n + 1)]
        
        dp[1][0] = 0
        dp[1][1] = 0
        dp[1][2] = 0

        for i in range(m):
            next_dp = [[-math.inf] * 3 for _ in range(n + 1)]
            for j in range(n):
                for k in range(3):
                    # Base transition (no neutralization here)
                    best_prev = max(dp[j + 1][k], next_dp[j][k])
                    if best_prev != -math.inf:
                        next_dp[j + 1][k] = max(next_dp[j + 1][k], best_prev + coins[i][j])
                    
                    # Transition with neutralization
                    if k > 0:
                        best_prev_power = max(dp[j + 1][k - 1], next_dp[j][k - 1])
                        if best_prev_power != -math.inf and coins[i][j] < 0:
                            next_dp[j + 1][k] = max(next_dp[j + 1][k], best_prev_power)
            
            # Special setup for first cell handling on first row
            if i == 0:
                next_dp[0] = [-math.inf] * 3
            dp = next_dp
            
        return max(dp[n])