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<vector>
#include<algorithm>
#include<string>
#include<queue>
#include<cstdio>
#include<cmath>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define ALL(x) x.begin(),x.end()
int MOD=1000000007;int inf=MOD*MOD;
int n,k;
int a[20];
int calc(vector<bool> const& status){
    int num=0;
    rep(i,n){
        if(status[i])num++;
    }
    if(num<k)return inf;
    int res=0;
    int last=a[0];
    rep1(i,n-1){
        if(status[i]){
            res+=max((int)0,last-a[i]+1);
            last=max(last+1,a[i]);
        }
    }
    return res;
}
int solve(vector<bool> status,int cur){
    if(cur==n)return calc(status);
    else{
        int res=solve(status,cur+1);
        status[cur]=true;
        res=min(res,solve(status,cur+1));
        return res;
    }
}
signed main(){
    cin>>n>>k;
    rep(i,n)cin>>a[i];
    vector<bool> status(n,false);
    status[0]=true;
    int res=solve(status,1);
    cout<<res<<endl;
    return 0;
}