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)

/*
    Problem 14
    https://atcoder.jp/contests/s8pc-4/tasks/s8pc_4_b
*/
#include <bits/stdc++.h>
using namespace std;
/* typedef */
typedef long long ll;
typedef pair<int, int> pii;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 50;
const int NIL = -1;
const int MAX = 10000;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */
/* function */
/* main */
int main(){
    int n, k;
    cin >> n >> k;
    vector<ll> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];

    ll ans = LINF;
    // 一番前は絶対見えるので
    // 2番目以降からK - 1個選んで見えるようにする
    for (int b = 0; b < (1 << (n - 1)); b++) {
        if (__builtin_popcount(b) != k - 1) continue;
        ll maxH = a[0], sum = 0;
        for (int i = 1; i < n; i++) {
            if ((b >> (i - 1)) & 1) {
                if (maxH >= a[i]) {  // 低ければmaxH+1まで増築
                    maxH += 1;
                    sum += maxH - a[i];
                }
            }
            maxH = max(a[i], maxH);  // 高さの更新
        }
        ans = min(sum, ans);
    }
    cout << ans << '\n';
}