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.BufferedWriter;
import java.io.FileWriter;
import java.io.InputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.util.Comparator.*;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ======================================================================
    public void solve() {
    	int N = ni();
    	int K = ni();
    	long[] A = ndl(N);
    	long ans = (long)(1e18), wk, sv, cnt;
    	boolean f;
    	for (int i = 0; i < (int)(1 << N); i++) {
    		if(Integer.bitCount(i) != K
    						&& (i & 0) == 0)	continue;
    		sv = A[0];
    		cnt = 1;
    		wk = 0;
    		for (int j = 1; j < N; j++) {
				if((i & (1 << j)) != 0) {
					if(A[j] <= sv) {
						wk += sv - A[j] + 1;
						sv = sv + 1;
					} else {
						sv = A[j];
					}
					cnt++;
				} else if(A[j] > sv) {
					wk += A[j] - sv;
					sv = A[j];
				} else if(A[j] == sv) {
					wk++;
					sv++;
				}
	    		if(cnt >= K)	break;
			}
    		ans = Math.min(ans, wk);
    	}
    	out.println(ans);
    }
    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}