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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("O3") //コンパイラ最適化用

#define _GLIBCXX_DEBUG //配列に[]でアクセス時のエラー表示
#include <algorithm>   //sort,二分探索,など
#include <bitset>      //固定長bit集合
#include <chrono>      //実行時間計測
#include <cmath>       //pow,logなど
#include <complex>     //複素数
#include <deque>       //両端アクセスのキュー
#include <functional>  //sortのgreater
#include <iomanip>     //setprecision(浮動小数点の出力の誤差)
#include <iostream>    //入出力
#include <iterator>    //集合演算(積集合,和集合,差集合など)
#include <map>         //map(辞書)
#include <numeric>     //iota(整数列の生成),gcdとlcm,accumulate
#include <queue>       //キュー
#include <set>         //集合（順序保持）
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility> //pair
#include <vector>  //可変長配列
using namespace std;
typedef long long ll;
typedef long double ld;

#define REP(i, n) for (int i = 0; i < ll(n); i++)
#define REPD(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = a; i < ll(b); i++)
#define FORD(i, a, b) for (int i = a; i > ll(b); i--)
// xにはvectorなどのコンテナ
#define ALL(x) x.begin(), x.end()
#define SIZE(x) ll(x.size()) // sizeをsize_tからllに直しておく
const long long INF = 1e12;
const int MOD = 1e9 + 7;
#define MAXR 100000 // 10^5:配列の最大のrange(素数列挙などで使用)
//略記
#define Umap unordered_map
#define Uset unordered_set
#define int long long
//#define int long long
const unsigned int BIT_FLAG_0 = (1 << 0); // 0000 0000 0000 0001
const unsigned int BIT_FLAG_1 = (1 << 1); // 0000 0000 0000 0010
const unsigned int BIT_FLAG_2 = (1 << 2); // 0000 0000 0000 0100
const unsigned int BIT_FLAG_3 = (1 << 3); // 0000 0000 0000 1000
const unsigned int BIT_FLAG_4 = (1 << 4); // 0000 0000 0001 0000
const unsigned int BIT_FLAG_5 = (1 << 5); // 0000 0000 0010 0000
const unsigned int BIT_FLAG_6 = (1 << 6); // 0000 0000 0100 0000
const unsigned int BIT_FLAG_7 = (1 << 7); // 0000 0000 1000 0000

// https://qiita.com/Lily0727K/items/06cb1d6da8a436369eed#c%E3%81%A7%E3%81%AE%E5%AE%9F%E8%A3%85
void print() { cout << endl; }

template <class Head, class... Tail> void print(Head &&head, Tail &&... tail) {
    cout << head;
    if (sizeof...(tail) != 0)
        cout << " ";
    print(forward<Tail>(tail)...);
}

template <class T> void print(vector<T> &vec) {
    for (auto &a : vec) {
        cout << a;
        if (&a != &vec.back())
            cout << " ";
    }
    cout << endl;
}

template <class T> void print(vector<vector<T>> &df) {
    for (auto &vec : df) {
        print(vec);
    }
}

void add(long long &a, long long b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
}
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

signed main() {

    int n, k;
    cin >> n >> k;
    vector<int> A(n);
    REP(i, n) { cin >> A.at(i); }

    if (n == k) {
        int height = A[0];
        int ans = 0;
        FOR(i, 1, n) {
            if (A[i] <= height) {
                ans += height - A[i] + 1;
                height += 1;

            } else {
                height = A[i];
            }
        }
        cout << ans << endl;
    } else {
        // https://drken1215.hatenablog.com/entry/2019/12/14/171657
        int minans = INF;
        // reverse(ALL(A));
        for (ll bit = 0; bit < (1 << n); ++bit) {
            if (__builtin_popcount(bit) != k)
                continue;
            int cnt = 0;
            int ans = 0;
            int height = 0;

            for (ll i = 0; i < n; ++i) {

                if (!(bit & (1 << i))) {
                    chmax(height, A[i]);
                } else {
                    if (i == 0) {
                        height = A[i];
                    } else {
                        if (A[i] <= height) {
                            ans += height - A[i] + 1;
                            height += 1;

                        } else {
                            height = A[i];
                        }
                        cnt++;
                    }
                }
                if (cnt == k) {
                    break;
                }
            }
            chmin(minans, ans);
        }
        print(minans);
    }
    return 0;
}