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: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();
		int K = sc.nextInt();

		int[] a = nextIntArray(sc, N);

		sc.close();

		long ans = Long.MAX_VALUE;

		for (int bit = 0; bit < (1 << N); bit++) {
			if (Integer.bitCount(bit) != K || (bit&1)!=1) {
				continue;
			}

			long t = a[0];
			long c = 0;
			for (int j = 1; j < N; j++) {
				if (((bit >> j) & 1) == 1) {
					c += Math.max( (t + 1) - a[j], 0);
					t = Math.max(t + 1, a[j]);
				} else {
					t = Math.max(t, a[j]);
				}
			}

			ans = Math.min(ans, c);
		}

		out.println(ans);

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}

	@SuppressWarnings("unused")
	private int[] nextIntArray(Scanner sc, int n) {
		int[] result = new int[n];
		for (int i = 0; i < n; i++) {
			result[i] = sc.nextInt();
		}
		return result;
	}
}