def max_beauty_sum(N, K, T):
    # dp[i] will store the max sum of the minimums when partitioning first i tulips into up to K bouquets
    dp = [[-float('inf')] * (K + 1) for _ in range(N + 1)]
  
    # Base case: 0 tulips, 0 bouquets has a sum of 0
    dp[0][0] = 0
  
    # Iterate over the number of bouquets
    for k in range(1, K + 1):
        # Monotonic stack to keep track of minimums efficiently
        stack = []
        for i in range(k, N + 1):
            current_min = float('inf')
            # Calculate dp[i][k] as we scan from i backwards
            for j in range(i, k - 1, -1):
                current_min = min(current_min, T[j - 1])
                # Update dp[i][k] using the previous state dp[j-1][k-1]
                dp[i][k] = max(dp[i][k], dp[j-1][k-1] + current_min)
  
    # The answer is dp[N][K] which stores the maximum possible sum
    return dp[N][K]
# Input parsing
N, K = map(int, input().split())
T = list(map(int, input().split()))
# Output the result
print(max_beauty_sum(N, K, T))