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)

// 左端は必ず見えるのでその他で見えるものを(n-1)C(k-1)通り列挙してシミュ

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) (x).begin(),(x).end()
using namespace std;
const int INF=1145141919,MOD=1e9+7;
const long long LINF=8931145141919364364,LMOD=998244353;
inline long long mod(long long n,long long m){return(n%m+m)%m;}
// const int dx[]={1,0,-1,0,1,1,-1,-1},dy[]={0,-1,0,1,1,-1,-1,1};

int main(){
    long long n,k; cin>>n>>k;
    long long front; cin>>front;
    vector<long long> a(n-1); rep(i,n-1) cin>>a[i];
    vector<int> v;
    rep(i,n-k) v.push_back(0);
    rep(i,k-1) v.push_back(1);
    long long ans=LINF;
    do{
        vector<long long> b=a;
        long long mini=front+1,sum=0;
        rep(i,n-1){
            if(v[i]){
                sum+=max(mini-b[i],0LL);
                b[i]=mini;
            }
            mini=max(mini,b[i]+1);
        }
        // cout<<sum<<endl;
        ans=min(ans,sum);
    }while(next_permutation(all(v)));
    cout<<ans<<endl;
    return 0;
}