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<iostream>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<iomanip>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define fi first
#define sc second
#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())

const int MAX = 510000;
const int MOD = 1000000007;

int main(){
  ll N, K; cin >> N >> K;
  vector<ll> a(N);
  rep(i,N) cin >> a[i];

  ll ans = 1e16;

  rep(i, (1 << N)){
    if( i % 2 == 0) continue;
    bool pre = false;
    ll kcnt = 0;
    rep(j , N){
      if((i >> j) % 2 == 1) kcnt++;
    }
    if(kcnt == K) pre = true;

    // preが成り立つときにcostcheckする
    if(pre){
      ll cost = 0;
      ll h = a[0];
      // hは現時点での高さ、これを順次超える必要あり
      rep(j ,N){
        if((i >> j) % 2 == 1){
          cost += max(h - a[j] , (ll)0);
          h = max(++h, a[j] + 1);
        }
      }
      // cout << i << " " << cost << endl;
      ans = min(ans , cost);
    }
  }
  cout << ans << endl;
}