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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int N;
  cin  >> N;
  vector<int64_t> v(N);
  for(int i=0;i<N;i++){
    cin >> v.at(i);
    v.at(i)-=(i+1);
  }
  int64_t sump=0;
  int64_t countp=0;
  int64_t summ=0;
  int64_t countm=0;
  int64_t countz=0;
  for(int i=0;i<N;i++){
    if(v.at(i)<0){
      summ-=v.at(i);
      countm++;
    }else if(v.at(i)>0){
      sump+=v.at(i);
      countp++;
    }else{
      countz++;
    }
  }
  int64_t ans=sump+summ;
  if(countp==countm){
    cout << ans << endl;
    return 0;
  }
  if(countm>countp){
    for(int i=0;i<N;i++){
      v.at(i)=-v.at(i);
    }
    swap(countp,countm);
  }
  sort(v.begin(),v.end());
  countm+=countz;
  int prev=0;
  int diff;
  for(int i=0;i<N;i++){
    if(v.at(i)>0){
      diff=v.at(i)-prev;
      sump-=diff*countp;
      summ+=diff*countm;
      ans=min(ans,sump+summ);
      countp--;
      countm++;
      prev=v.at(i);
    }
  }
  cout << ans << endl;
  return 0;
}