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>
#include <numeric>
#include <limits>
#include <cassert>
#include <cstdint>
using namespace std;
int64_t solve(vector<int64_t> &as, int64_t n)
{
    for(int64_t i = 0; i < as.size(); ++i)
    {
        as[i] -= i+1;
    }
    sort(as.begin(), as.end());

    int64_t lsum = 0;
    int64_t rsum = accumulate(as.begin(), as.end(), 0);
    int64_t value = numeric_limits<int64_t>::max();
    //cout << "lsum:" << lsum << ", rsum: " << rsum << endl;
    for(int64_t i = 0; i < as.size(); ++i)
    {
        lsum -= as[i];
        rsum -= as[i];
        //cout << "as[" << i << "]: " << as[i] << endl;
        //cout << "lsum:" << lsum << ", rsum: " << rsum << endl;        
        value     = min<int64_t>(lsum + rsum + ( 2 * (i+1) - n) * as[i  ], value);
        if ( i < as.size() - 1)
        {
            value = min<int64_t>(lsum + rsum + ( 2 * (i+1) - n) * as[i+1], value);    
        }
        //cout << value << endl;
        //assert(value >= static_cast<int64_t>(0));
    }
    return value;
}
int main()
{
    int n;
    cin >> n;
    vector<int64_t> as;
    for(int i = 0; i < n; ++i)
    {
        int a;
        cin >> a;
        as.push_back(a);
    }
    cout << solve(as, n) << endl;
}