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>
#include <math.h>
#include <cmath>

using namespace std;
using ll =long long;
using vi = vector<int>; 
using vvi = vector<vi>; 
using vl = vector<ll>; 
using vvl = vector<vl>; 
using vs = vector<string>;
using vvs = vector<vs>; 
using vc = vector<char>;
using vvc = vector<vc>; 
using vb = vector<bool>;
using vvb = vector<vb>; 

#define rep(i,n); for( int i =0; i < n ; i ++)
#define all(a) a.begin(),a.end()
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
#define pri(a)  printf("%.14lf\n",a);
#define MOD 1000000007
typedef pair<int, int> P;
bool is_int_lround(double x){
  return std::lround(x)==x;
}
ll keta(ll x){
  ll n=0;
  while(x>0){
    x /=10 ;
    n ++;
  }
  return n;
}

ll conbi(int n,int m){
  
  cin>>n>>m;
  vector<ll> a(100);
  a[0] =1;
  for(int i=0;i<14;i++){
    a[i+1]=a[i]*(i+1);
  }
  return a[n] /(a[m] *a[n-m]);
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;//繰り返し二乗法
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
ll kaijo(ll x){
  ll z=1;
  while(x>0){
    z *=x;
    z %=MOD;
    x--;
  }
  return z;
}
ll yakusu_num(ll n){
  vl yakusu(n+1,1);
  for(ll i=2;i*i<=n;i++){ 
    while(n%i==0){
      n /=i;
      yakusu[i]++;
    }
  }
  if(n!=1)yakusu[n]++;
  ll num=1;
  for(ll i=0;i <=n;i++){
    num*=yakusu[i];
  }
  return num;
}


//cout<<""<<endl;
int main(){
  ll N,K;cin>>N>>K;vl a(N);
  rep(i,N){
    cin>>a[i];
  }

  ll ans=10000000000;
  for(int tmp ; tmp<(1<<(N-1));tmp++){
    bitset<14> s(tmp);
    ll cnt=0;
    ll ok=1;
    ll ma=a[0];
    for(int i=0;i<N-1;i++){
      if(a[i+1]>ma)ok++;
      if(s.test(i)){
        if(ma<a[i+1])ma=a[i+1];
        else{
          cnt+=-a[i+1]+ma+1;
          ma++;ok++;
        }
      }
      else ma =max(ma,a[i+1]); 
    }
    if(ok>=K)ans=min(ans,cnt);
  }
  cout<<ans<<endl;
}