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<map>
#include<set>
#include<queue>
#include<deque>
#include<iomanip>
#include<sstream>
#include<cmath>
#include<bitset>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define req(i,n) for(long long int i = 1;i <=  n; i++)
#define rrep(i,n) for(int i = n -1;i >= 0;i--)
#define ALL(obj) begin(obj), end(obj)
typedef long long int ll;
typedef long double ld;
const ll INF = (1 << 30);
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T & val) {
    fill((T*)array, (T*)(array + N), val);
}
int main() {
    ll n, k,a[15],sum,ans = INF; cin >> n >> k;
    rep(i, n) cin >> a[i];
    n--; k--;
    rep(i, (1 << n)) {
        int cnt = 0;
        rep(j, n) {
            if (i & (1 << j))cnt++;
        }
        if (cnt >= k) {
            sum = 0; ll fm = a[0];
            rep(l, n) {
                if (i & (1 << l)) {
                    sum += max(0LL, fm - a[l + 1] + 1);
                    fm = max(fm, a[l + 1] + max(0LL, fm - a[l + 1] + 1));
                }
                else {
                    fm = max(fm, a[l + 1]);
                }
            }
            ans = min(ans, sum);
        }
    }cout << ans << endl;
}