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>

int abs(int a)
{
    return (a > 0) ? a : -a;
}

void sort(int n, int B[n])
{
    int i, j;
    for(i = 0; i < n - 1; i++){
        for(j = n - 2; j > i; j--){
            if(B[j + 1] > B[j]){
                int tmp;
                tmp = B[j + 1];
                B[j + 1] = B[j];
                B[j] = tmp;
            }
        }
    }
}

int main()
{
    int n;  scanf("%d", &n);
    int A[n];


    int i;
    int B[n];
    for(i = 0; i < n; i++){
        scanf("%d", &A[i]);
        B[i] = A[i] - (i + 1);
    }

    sort(n, B);

    if(n % 2 == 0){
        int b1, b2;
        b1 = B[n / 2];
        b2 = B[n / 2 - 1];
        int sum1 = 0, sum2 = 0;
        for(i = 0; i < n; i++){
            sum1 += abs(B[i] - b1);
            sum2 += abs(B[i] - b2);
        }
        printf("%d\n", (sum1 < sum2) ? sum1 : sum2);
    }
    else{
        int b;
        int sum = 0;
        b = B[n / 2 + 1];
        for(i = 0; i < n; i++){
            sum += abs(B[i] - b);
        }
        printf("%d\n", sum);
    }

    return 0;
}