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: Wrong Answer)

use std::str::FromStr;

fn get_one<T : FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().parse().ok().unwrap();
}

fn get_vec<T : FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();
}

fn get_vec_by_lines<T : FromStr>(n : usize) -> Vec<T> {
    let mut v : Vec<T> = Vec::new();
    for _i in 0..n {
        let x : T = get_one();
        v.push(x);
    }
    return v;
}


fn main() {
    let n : usize = get_one();
    let mut v : Vec<i32> = get_vec();
    for i in 1..n+1 {
        v[i-1] -= i as i32;
    }
    v.sort();
    let m = v[0];
    for i in 0..n {
        v[i] -= m;
    }
    let sm = v.iter().fold(0, |s, x| s + x);
    //println!("{}", sm);
    let mut cur = sm;
    let mut ans = sm;
    for i in 1..n {
        cur = cur - (v[i] - v[i-1]) * ((n-i) as i32) + (v[i]-v[i-1]) * (i as i32);
        //println!("{} {} {} {} {}", i, v[i], v[i-1], n-i, cur);
        ans = std::cmp::min(ans,cur.abs());
    }

    println!("{}", ans);
}