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<vector>

using namespace std;

int main(){
  int N;
  cin >> N;
  
  vector<int> A(N);
  int amin = 100000;
  int imin = 100000;
  for(int i = 0; i < N; i++){
    cin >> A[i];
    if(A[i] <= amin) imin = i;
    amin = min(A[i],amin);
  }
  int bmin = amin - imin;
  //総和最小値を取りうるbの最小値=Aminを通る直線y=x+bの切片b b = Amin - imin
  //b=最小値から一個づつ増加させてgreedy(残差の減少がなくなった点が最小）;
  int d = 0;
  int lastd = 0;
  int ans = 0;
  int f = 0;
  for(int i = bmin; i < 100000; i++){
    for(int j = 0; j < N; j++) d += abs(A[j] - (j + 2));
    if(lastd <= d){
      ans = lastd;
      f = 1;
      break;
    }
    if(f) break;
    lastd = d;
    d = 0;
  }
  cout << ans;
}