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 (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n = read::<usize>();
    let mut aseq = vec![0; n];
    for i in 0..n {
        aseq[i] = read::<i64>();
    }
    let mut sum = 0;
    let b = -1e+9 as i64;
    let mut ds = vec![0; n];
    let mut cnt = n as i64;
    for i in 0..n {
        sum += (aseq[i] - (b + i as i64 + 1)).abs();
        ds[i] = aseq[i] - (b + i as i64 + 1);
    }
    assert!(sum >= 0);
    ds.sort();
    let mut ans = sum;
    let mut prev = 0;
    // println!("{}", sum);
    for i in 0..n {
        let d = ds[i] - prev;
        sum -= d * cnt;
        // println!("{} {} {}", sum, d, cnt);
        cnt -= 2;
        prev = ds[i];
        ans = std::cmp::min(ans, sum);
        assert!(sum >= 0);
    }
    println!("{}", ans);
}