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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		int n = Integer.parseInt(str[0]);
		int k = Integer.parseInt(str[1]);

		long a[] = new long[n];

		str = br.readLine().split(" ");
		for (int i = 0; i < n; i++) {
			a[i] = Long.parseLong(str[i]);
		}

		long ans = Long.MAX_VALUE;
		for (int bit = 0; bit < (1 << n); bit++) {
			long b[] = a.clone();
			boolean flag[] = new boolean[n];
			flag[0] = true;

			int count = 0;
			for (int i = 0; i < n; i++) {
				if ((bit & (1 << i)) != 0) {
					flag[i] = true;
					count++;
				}
			}

			if (count < k)
				continue;

			long sum = 0;
			for (int i = 0; i < n; i++) {
				long height = 0;
				if (flag[i]) {
					for (int j = 0; j < i; j++) {
						if (flag[j]) {
							height = b[j];
						}
					}
					if (height >= b[i]) {
						sum += Math.max(0, height + 1 - b[i]);
						b[i] += Math.max(0, height + 1 - b[i]);
					}
				}
			}
			ans = Math.min(ans, sum);
		}
		System.out.println(ans);
	}
}