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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int Inf = 1000000000;
constexpr ll INF= 1e18;
constexpr ll MOD = 1000000007;
const double PI = 3.1415926535897;
typedef pair<ll,ll> P;

int main() {
    int N,K;
    cin >> N >> K;
    vector<ll> vec(N);
    for(int i = 0;i < N;i++) {
        cin >> vec.at(i);
    }
    ll ret = INF;
    for(int bit = 0;bit < (1 << N);bit++) {
        vector<ll> cnt;
        for(int i = 0;i < N;i++) {
            if(bit & (1 << i)) {
                cnt.push_back(vec.at(i + 1));
            }
        }
        if(cnt.size() >= K) {
            ll ret2 = 0;
            for(int i = 0;i < cnt.size() - 1;i++) {
                ret2 += max(0ll,cnt.at(i) + 1 - cnt.at(i + 1));
                cnt.at(i + 1) = max(cnt.at(i + 1),cnt.at(i) + 1);
            }
            ret = min(ret,ret2);
        }
    }
    cout << ret << endl;
}