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: Wrong Answer)

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


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        var n = long.Parse(Console.ReadLine());
        long[] a = Console.ReadLine().Split().Select(long.Parse).ToArray();

        long[] abs = new long[n];
        for (int i = 0; i < n; i++)
        {
            abs[i] = a[i] - (i + 1);
        }

        // abs配列要素の登場回数をカウント
        var count = new Dictionary<long, long>() { };
        for (int i = 0; i < n; i++)
        {
            if (!count.ContainsKey(abs[i]))
            {
                count.Add(abs[i], 1);
            }
            else
            {
                count[abs[i]]++;
            }
        }

        var tmp = count.OrderByDescending(val => val.Value).First().Value;
        long b = long.MaxValue;
        foreach (var item in count)
        {
            if (item.Value == tmp)
            {
                if (b > Math.Abs(item.Key))
                {
                    b = item.Key;
                }
            }
        }

        long ans = 0;
        for (int i = 0; i < n; i++)
        {
            ans += Math.Abs(a[i] - (b + i + 1));
        }


        // 出力
        Console.WriteLine(ans);
    }
}