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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <type_traits>
#include <map>
#include <limits>

using namespace std;


int main(int argc, const char * argv[]) {
  long long n;
  cin >> n;
  double sum1 = 0;
  double sum2 = 0;

  vector<int> a(n);
  vector<int> bs(n);
  for(int i = 0; i < n; i++){
    cin >> a[i];
    bs[i] = (a[i] - (i + 1));
  }
  sort(bs.begin(), bs.end());
//  cout << bs[n/2] << endl;
  for(int i = 0; i < n; i++){
    sum1 += abs(a[i] - (bs[n/2] + i + 1 ));
    sum2 += abs(a[i] - (bs[n/2-1] + i + 1 ));
  }
  cout << min(sum1, sum2) << endl;
  return 0;
}