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 <bits/stdc++.h>
using namespace std;
 
#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >=0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)
 
typedef long long ll;
#define INF 1e9


int main(){
  int N;
  cin >> N;
  vector<int> A(N), sum(N+1);
  REP(i,N){
    cin >> A[i];
    A[i] -= i+1;
    sum[i+1] = sum[i] + A[i];
  }
  
  int ave = sum[N] / N;
  //cout << ave << endl;
  int mins = 1e9;
  ll sums = 0;
  REP(i, N){
    sums += abs(A[i] - ave);
  }
  mins = sums;
    ave++;
    sums = 0;
    REP(i, N){
      sums += abs(A[i] - ave);
    }
    mins = min((ll)mins, sums);

  ave -=2;
  //cout << ave << endl;

  sums = 0;
    REP(i, N){
      sums += abs(A[i] - ave);
    }
    mins = min((ll)mins, sums);
  cout << mins << endl;

  return 0;
}