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>
#include<algorithm>
using namespace std;


int main() {
    int n;
    cin>>n;
    vector<int> a(n), tmp(n);
    long long sum = 0;
    for(int i=0; i<n; i++) {
        cin>>a[i];
        tmp[i]=a[i];
    }
    if(n%2==1) {
        long long ave = tmp[n/2] - (n+1)/2;
        long long sad1=0, sad2=0;
        for(int i=0; i<n; i++) {
            sad1 += abs(a[i]-ave-i-1);
            sad2 += abs(a[i]-ave-i);
        }
        cout<<min(sad1, sad2)<<endl;
    } else{
        long long ave = (tmp[n/2-1]+tmp[n/2])/2 - (n+1)/2;
        long long sad1=0, sad2=0;
        for(int i=0; i<n; i++) {
            sad1 += abs(a[i]-ave-i-1);
            sad2 += abs(a[i]-ave-i);
        }
        cout<<min(sad1, sad2)<<endl;
    }

}