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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

namespace Practice
{
    class Program
    {

        static void Main(string[] args)
        {
            //SolveA(args);
            //SolveB(args);
            SolveC(args);
            //SolveD(args);
        }
        
        static void SolveA(string[] args)
        {
            string sz = (Console.ReadLine());
            long n = long.Parse(sz);
            if(n %  2 == 0)
            {
                Console.WriteLine(n);
            }
            else
            {
                Console.WriteLine(n * 2);
            }
        }
        static void SolveB(string[] args)
        {
            string sz = (Console.ReadLine());
            long n = long.Parse(sz);
            long[] a = new long[n];
            string[] szArr = Console.ReadLine().Split(' ');

            long max = long.MinValue;
            long min = long.MaxValue;

            for (int i=0;i<szArr.Length;++i)
            {
                long tmp  = long.Parse(szArr[i]);
                a[i] = tmp;
                max = Math.Max(tmp, max);
                min = Math.Min(tmp, min);
            }

            long val = Math.Abs(max - min);
            Console.WriteLine(val);
        }
        static void SolveC(string[] args)
        {
            string sz = (Console.ReadLine());
            long n = long.Parse(sz);
            long[] a = new long[n];
            string[] szArr = Console.ReadLine().Split(' ');
            for (int i = 0; i < szArr.Length; ++i)
            {
                long tmp = long.Parse(szArr[i]);
                a[i] = tmp - (i + 1);

            }


            long[] a2 = a.OrderByDescending(p => p).ToArray();
            long b_candidate = 0;
            long b_candidate2 = 0;

            if (n % 2 == 1)
            {
                long idx = (n - 1) / 2;
                b_candidate = a2[idx];
            }
            else
            {
                long k = (n-1) / 2;
                b_candidate = a2[k];
                b_candidate2 = a[k + 1];
            }
            var sad_list = new List<long>();
            sad_list.Add(SadValue(a, b_candidate));
            sad_list.Add(SadValue(a, b_candidate2));

            long sad = sad_list.Min();
            Console.WriteLine(sad);

        }
        private static double Average(long[] a)
        {
            // オーバーフローしそう

            double ave = 0.0;
            long total = 0;
            for(int i=0;i<a.Length;++i)
            {
                ave += (a[i] - ave) / (total + 1);
                total++;
            }
            return ave;
        }
        private static long SadValue(long[] a, long b)
        {
            long val = 0;
            for(int i=0;i<a.Length;++i)
            {
                val += Math.Abs(a[i] - b);
            }
            return val;
        }
    }

}