def max_sum_with_breaks(arr, k):
    n = len(arr)
    dp = [-float('inf')] * n  # Initialize dp with -inf to find the maximum
    dp[0] = arr[0]  # Base case: first element can only be taken by itself

    for i in range(1, n):
        current_sum = 0
        # Consider taking up to k successive elements ending at index i
        for j in range(k):
            if i - j >= 0:  # Ensure we're within bounds
                current_sum += arr[i - j]  # Add the element arr[i-j]
                if i - j - 2 >= 0:
                    dp[i] = max(dp[i], dp[i - j - 2] + current_sum)  # Consider the break
                else:
                    dp[i] = max(dp[i], current_sum)  # No previous group, just the current sum

    return max(dp)

# Test cases:
list1 = [60, 70, 80, 40, 80, 90, 100, 20]
k1 = 3
print(max_sum_with_breaks(list1, k1))  # Expected output: 480

list2 = [45, 12, 78, 34, 56, 89, 23, 67, 91]
k2 = 4
print(max_sum_with_breaks(list2, k2))  # Expected output: 460