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::io::stdin;

fn abs(val: i32) -> i32 {
    if val >= 0 { val } else { -val }
}

/*
∑|Ai - (b + i)|

4  6  2  3  6  6  7  9
1  2  3  4  5  6  7  8

 3  4 -1 -1  1  0  0  1
-1 -1  0  0  1  1  3  4
 1  2  3  4  5  6  7  8
          ^  ^

Ai - b - i
4  - X - 1
X = 3

b = 3
0 1 4 4 2 3 3 2 = 19

6  - X - 2
X = 4

b = 4
1 0 5 5 3 4 4 3 = 25

b = -1
4 5 0 0 2 1 1 2 = 15

b = 0
3 4 1 1 1 0 0 1 = 11

b = 1
2 3 2 2 0 1 1 0 =
*/

fn main() {
    let mut n = String::new();
    stdin().read_line(&mut n);
    let n: i64 = n.trim().parse().unwrap();

    let mut a = String::new();
    stdin().read_line(&mut a);
    let A: Vec<i32> = a.trim().split_whitespace().map(|i|i.parse().unwrap()).collect();
    let mut B: Vec<i32> = A.iter().enumerate().map(|(i, &a)| a - 1 - i as i32).collect();

    let f = |b: i32| {
        A.iter().enumerate().fold(0i32, |acc, (i, &a)| acc + abs(a - (b + (i as i32 + 1))))
    };
//    B.sort_by(|&a, &b| f(a).cmp(&f(b)));
    B.sort();
    if B.len() % 2 == 1 {
        println!("{}", f(B[B.len() / 2]));
    } else {
        println!("{}", f((B[B.len() / 2] + B[B.len() / 2 - 1]) / 2));
    }
//    println!("{}", B.iter().fold(1 << 30, |acc, &b| {
//        let s = f(b);
//        if acc > s { s } else { acc }
//    }));
}