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<stdio.h>
#include<math.h>
long long a[20];
long long sum;
int n,m;
void dfs(int x,int y,int z,long long t)
{
    int i,j;
    if(y==m)
    {
        if(t<sum)
        sum=t;
        return;
    }
    for(i=x+1;i<n;i++)
    {
        if(a[i]<=z)
        {
        j=z+1;
        dfs(i,y+1,j,t+z-a[i]+1);
        }
        else
        {
        j=a[i];
        dfs(i,y+1,j,t);
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        sum=1000000000000000;
            dfs(0,1,a[0],0);
        printf("%lld\n",sum);
    }
        return 0;
}