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 (1.42.0) (judge verdict: Accepted)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn main() {
    input! {
        n: usize,
        k: usize,
        a: [i64; n],
    }

    let mut ans = std::i64::MAX;
    for bit_row in 0..2u32.pow(n as u32) {
        // selected の建物で条件を満たす
        let mut selected = vec![];
        for i in 0..n {
            if bit_row & (1 << i) > 0 {
                selected.push(i);
            }
        }
        if selected.len() != k {
            continue;
        }
        // println!("{:?}", selected);

        let mut current = 0;
        let mut required = 0;
        let mut selected_idx = 0;
        for i in 0..n {
            if i == selected[selected_idx] {
                // println!("required: {}", required);
                if required > a[i] {
                    current += required - a[i];
                    required = required + 1;
                } else {
                    required = a[i] + 1;
                }
                selected_idx += 1;
                if selected_idx == k {
                    break;
                }
            } else {
                required = required.max(a[i] + 1);
            }
        }
        // println!("{}", current);
        ans = ans.min(current);
    }
    println!("{}", ans);
}