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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(void){
    // Your code here!
    int key = 0, flag = 0, A[200000], N, max = 0, min = 1000000000;
    long sumleft, sumcandi, sumright, candi = 0, Min = LONG_MAX;
    scanf("%d",&N);
    for(int i = 1; i <= N; i++)
    {
        scanf("%d",&A[i-1]);
        A[i-1] = A[i-1] - i;
        
        if(A[i-1] > max) max = A[i-1];
        if(A[i-1] < min) min = A[i-1];
        
        candi += A[i-1];
    }
    
    candi /= N;
    
    while(flag == 0)
    {
        switch(key)
        {
            case 0:
                sumleft = sumcandi = sumright = 0;
                break;
            case 1:
                sumright = 0;
                break;
            case -1:
                sumleft = 0;
                break;
        }
        
        for(int j = 0; j < N; j++)
        {
            switch(key)
            {
                case 0:
                    sumcandi += labs(A[j] - candi);
                    sumleft += labs(A[j] - candi - 1);
                    sumright += labs(A[j] - candi + 1);
                    break;
                    
                case 1:
                    sumright += labs(A[j] - candi + 1);
                    break;
                    
                case -1:
                    sumleft += labs(A[j] - candi - 1);
                    break;
                    
            }
        }
        //printf("%ld %ld %ld\n", sumleft, sumcandi, sumright);
        
        if(sumcandi <= sumright && sumleft <= sumcandi)
        {
            if(key == 1)
            {
                Min = sumcandi;
                flag = 1;
            }
            else
            {
                candi--;
                key = -1;
                sumright = sumcandi;
                sumcandi = sumleft;
            }
        }
        else if(sumcandi >= sumright && sumleft >= sumcandi)
        {
            if(key == -1)
            {
                Min = sumcandi;
                flag = 1;
            }
            else
            {
                candi++;
                key = 1;
                sumleft = sumcandi;
                sumcandi = sumright;
            }
        }
        else
        {
            flag = 1;
            Min = sumcandi;
        }
    }
    
    printf("%ld\n",Min);
}