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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        
        int n_bld = Integer.parseInt(sc.next());
        int wanted_colors = Integer.parseInt(sc.next());
        int[] h_each = new int[n_bld];
        for(int i=0; i<n_bld; i++){
            h_each[i] = Integer.parseInt(sc.next());
        }
        int visible_n=0;
        int cost=0, min=Integer.MAX_VALUE;
        //int[] h_new = new int[n_bld];
        int highest_former=0;
        for(int bit=0; bit < (1<<n_bld); bit++){
            visible_n=0;
            highest_former=0;
            cost=0;
            for(int i=0; i<n_bld; i++){
                if(h_each[i] > highest_former){
                    highest_former=h_each[i];
                    visible_n++;
                }else if((bit&(1<<i)) > 0){
                    visible_n++;
                    highest_former++;
                    cost += (highest_former - h_each[i]);
                }
            }
            if((visible_n >= wanted_colors) && (cost < min)) min=cost;
        }
        
        
        System.out.println(min);
    }
}