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 <math.h>

main(){
    int n;
    scanf("%d",&n);
    int i;
    int a[n];
    long long ave=0;
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
        ave+=a[i]-i-1;
    }
    long long c,f;
    c=f=ave/n;
    if(ave>0)
        c++;
    else
        f--;
    long long k=0,l=0;
    ave=c;
    for(i=0;i<n;i++)
        k+=(a[i]-ave-i-1)>0?(a[i]-ave-i-1):-1*(a[i]-ave-i-1);
    ave=f;
    for(i=0;i<n;i++)
        l+=(a[i]-ave-i-1)>0?(a[i]-ave-i-1):-1*(a[i]-ave-i-1);
    if(k<l)
        printf("%lld\n",k);
    else
        printf("%lld\n",l);
    return 0;
}