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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include<stdbool.h>
typedef long long ll;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define REP(i,l,r) for(int i=l; i<(int)(r); i++)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
const ll MOD = 1000000007;

ll a[12];

int main(){
    int N,K; scanf("%d%d", &N, &K);
    rep(i,N){
        scanf("%d", a+i);
    }

    ll ans = MOD*MOD;
    rep(bit, (1 << N)) {
        if(__builtin_popcount(bit) < K) continue;
        ll aa[12];
        memcpy(aa, a, sizeof(a));
        ll tmp_a = 0, tmp_h = 0;

        rep(i, N) {
            if(bit&(1<<i)){
                if(aa[i] <= tmp_h){
                    tmp_a += tmp_h-aa[i]+1;
                    tmp_h++;
                }else{
                    tmp_h = aa[i];
                }
            }else{
                tmp_h = max(tmp_h,aa[i]);
            }
        }
        
        ans = min(ans, tmp_a);
    }
    
    printf("%lld\n", ans);
    return 0;
}