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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    readLine()
    val nums = readLine()!!.split(" ").map { it.toInt() }
    val nums_reduced = nums.mapIndexed { index, i -> i - index }

    var top = nums_reduced.max()!!
    var bottom = nums_reduced.min()!!
    var mid = (top + bottom) / 2
    while (true){
        if (betterToInclement(nums_reduced, mid)) {
            bottom = mid
            mid = (top + bottom)
        } else if (betterToDeclement(nums_reduced, mid)) {
            top = mid
            mid = (top + bottom)
        } else {
            break
        }
    }
    nums_reduced.sumBy { abs(it - mid) }.let { println(it) }

}

fun abs(a: Int) = if (a > 0) a else -a

fun betterToInclement(array: Iterable<Int>, b: Int) = array.sumBy { if (it > b) 1 else -1 } > 1
fun betterToDeclement(array: Iterable<Int>, b: Int) = array.sumBy { if (it < b) 1 else -1 } > 1