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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const int MOD2 = 998244353;

ll a[20];

int main(){
    int n, k;
    cin >> n >> k;
    rep(i, n) cin >> a[i];
    ll ans = LINF;
    for(int bits = 0; bits < (1<<n); bits++){
        if (bits&1 == 0 || __builtin_popcount(bits) != k) continue;
        ll now = a[0], tmp = 0;
        for(int i = 1; i < n; i++){
            if (bits&(1<<i)){
                if (now >= a[i]) {
                    now++;
                    tmp += now - a[i];
                }
                else now = a[i];
            }
        }
        ans = min(ans, tmp);
    }

    cout << ans << endl;

    return 0;
}