Problem:
Score : 300 points

Problem Statement

Snuke has an integer sequence A of length N.

He will freely choose an integer b.
Here, he will get sad if A_i and b+i are far from each other.
More specifically, the sadness of Snuke is calculated as follows:

abs(A_1 - (b+1)) + abs(A_2 - (b+2)) + ... + abs(A_N - (b+N))

Here, abs(x) is a function that returns the absolute value of x.

Find the minimum possible sadness of Snuke.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible sadness of Snuke.

Sample Input 1

5
2 2 3 5 5

Sample Output 1

2

If we choose b=0, the sadness of Snuke would be abs(2-(0+1))+abs(2-(0+2))+abs(3-(0+3))+abs(5-(0+4))+abs(5-(0+5))=2.
Any choice of b does not make the sadness of Snuke less than 2, so the answer is 2.

Sample Input 2

9
1 2 3 4 5 6 7 8 9

Sample Output 2

0

Sample Input 3

6
6 5 4 3 2 1

Sample Output 3

18

Sample Input 4

7
1 1 1 1 2 3 4

Sample Output 4

6

Solution in Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use proconio::{input, fastout};
use proconio::marker::*;
use nalgebra::min;

#[fastout]
fn main() {
    input! {
    n: usize,
    a: [i64; n]
    }
    let (mut l, mut r) = (-1_000_000_005, 1_000_000_005);
    let try_count = 100;
    for _ in 0..try_count{
        let mid_l = (l * 2 + r) / 3;
        let mid_r = (l + r * 2) / 3;
        if sadness(&a, mid_l) > sadness(&a, mid_r){
            l = mid_l;
        } else {
            r = mid_r;
        }
    }
    let mut ans = std::i64::MAX;
    for i in l..=r {
        ans = min(ans, sadness(&a, i));
    }
    println!("{}", ans);
}

fn sadness(a: &Vec<i64>, b: i64) -> i64 {
    a.iter().enumerate().map(|(i, &x)| (x - (b + i as i64 + 1)).abs()).sum::<i64>()
}