Problem:
Max Score: 350 Points

Problem Statement

There are N buildings along the line. The i-th building from the left is colored in color i, and its height is currently a_i meters.

Chokudai is a mayor of the city, and he loves colorful thigs. And now he wants to see at least K buildings from the left.

You can increase height of buildings, but it costs 1 yens to increase 1 meters. It means you cannot make building that height is not integer.

You cannot decrease height of buildings.

Calculate the minimum cost of satisfying Chokudai's objective.

Note: "Building i can see from the left" means there are no j exists that (height of building j) ≥ (height of building i) and j < i.

Input Format

N K
a_1 a_2 a_3 ... a_N

Output Format

Print the minimum cost in one line. In the end put a line break.

Constraints

1 ≤ K ≤ N ≤ 15

1 ≤ a_i ≤ 10^9

Scoring

Subtask 1 [120 points]

N = K

Subtask 2 [90 points]

N ≤ 5

a_i ≤ 7

Subtask 3 [140 points]

There are no additional constraints.

Sample Input 1

5 5
3949 3774 3598 3469 3424

Sample Output 1

1541

The optimal solution is (height of buildings from the left) = [3949, 3950, 3951, 3952, 3953].

Sample Input 2

5 3
7 4 2 6 4

Sample Output 2

7

The optimal solution is (height of buildings from the left) = [7, 8, 2, 9, 4].

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)

import copy
import sys

N, K = map(int, input().split())
A = list(map(int, input().split()))

def dfs(bitstring):
  if len(bitstring) == N:
    if sum(bitstring) == K:
      cost = 0
      B = copy.deepcopy(A)
      for i in range(1, N):
        if bitstring[i]:
          previousMax = max(B[0:i])
          if previousMax < B[i]:
            continue
          else:
            cost += previousMax - B[i] + 1
            B[i] = previousMax + 1
      return cost
    else:
      return sys.maxsize
  else:
    return min(dfs(bitstring + [False]), dfs(bitstring + [True]))

print(dfs([1]))