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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 1<<30;
const ll mod = 1000000007LL;

int next_combination(int sub) {//組み合わせ全列挙に使うやつ
    int x = sub & -sub, y = sub + x;
    return (((sub & ~y) / x) >> 1) | y;
}

int main(){
  ll N,K;
  cin>>N>>K;
  vector<ll> v(N);
  for(ll i = 0; i < N; i++) cin>>v[i];
  ll bit = (1<<K) -1;
  ll ans = INF;
  for(;bit<(1<<N);bit=next_combination(bit)){
    ll h = 0;
    ll cnt = 0;
    for(ll i = 0; i < N; i++){
      if(bit&(1<<i)) {
        if(h>=v[i]) {
          cnt+=h-v[i]+1;
          h++;
        }
        else chmax(h,v[i]);
      }
      else chmax(h,v[i]);
    }
    chmin(ans,cnt);
  }
  cout<<ans<<endl;
}