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 C# (.NET Core 3.1.201) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int num = int.Parse(Console.ReadLine());

            //入力した値を数値変換する
            string str = Console.ReadLine();
            string[] strArray = new string[num];
            strArray = str.Split(' ');
            int[] intArray = Array.ConvertAll(strArray, int.Parse);

            
            //for文を使用して悲しみの距離を計測する
            for (int i = 0; i < num; i++)
            {
                intArray[i] = intArray[i] - i-1;

                //負の値なら-1を掛ける
                if(intArray[i]<0)
                {
                    intArray[i] = intArray[i] * -1;
                }
            }

            //中央値
            int intMid = 0;
            if(num%2==1)
            {
                intMid= intArray [num/ 2 + 1] ;
            }
            else
            {
                intMid = (intArray[num / 2] + intArray[num / 2 + 1]) / 2;
            }

            //for文を使用して悲しみの距離を計測する
            for (int i = 0; i < num; i++)
            {
                intArray[i] = intArray[i] - intMid;

                //負の値なら-1を掛ける
                if (intArray[i] < 0)
                {
                    intArray[i] = intArray[i] * -1;
                }
            }


            //配列の中身をすべて合計する
            int intSum = 0;
            for(int i=0;i< num; i++)
            {
                intSum = intSum + intArray[i];
            }

            Console.WriteLine(intSum);

        }
    }
}