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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::*;

#[allow(dead_code)]
fn read<T: str::FromStr>() -> T {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf.trim().parse().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T: str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|a| a.parse().ok().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_vec2d<T: str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

// use std::collections::HashSet;
// use std::collections::HashMap;

fn main() {
    let v: Vec<usize> = read_vec();
    let (n, k) = (v[0], v[1]);
    let b: Vec<usize> = read_vec();
    let first_b = b[0];
    let mut cost = 0;
    //ASSUME n == k
    let mut prev = first_b;
    for i in 1..b.len() {
        if prev < b[i] {
            prev = b[i];
            continue;
        }
        prev += 1;
        cost += prev - b[i];
    }
    println!("{}", cost);
}