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: Wrong Answer)

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

int main(){
  long long N;
  cin>>N;
  long long A[N];
  long long sum=0;
  long long div=0;
  for(int i=0;i<N;i++){
    cin>>A[i];
    sum+=A[i];
    div+=(i+1);
  }
  //cout<<sum<<" "<<div<<" "<<N<<endl;
  long long T=(sum-div)/N;
  long long ans=0;
  long long minans=1000000000;
  for(int i=T-1;i<=T+1;i++){
    ans=0;
    for(int j=0;j<N;j++){
      ans+=abs(A[j]-(j+i));
      //cout<<abs(A[j]-(j+i));
    }
    //cout<<i<<" "<<ans<<"#"<<endl;
    minans=min(ans,minans);
  }
  cout<<minans<<endl;
  return 0;
}