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 main()
{
    int a[200000],n,x,y,z,sum=0,an,mi=0,pl=0,sortmi,sortpl,cou=0;
    scanf("%d",&n);

    for(x=0;x<n;x++){
        scanf("%d",&a[x]);
        a[x]-=x+1;
        if(a[x]>=0)
            pl++;
        if(a[x]<0)
            mi++;
    }
    for(x=0;x<n;x++){
        if(a[x]>0)
            sum+=a[x];
        if(a[x]<0)
            sum+=a[x]*-1;
    }
    an=sum;
    sum=0;

    if(pl-mi>0){
        for(x=0;x<1000000000;x++){
            for(y=0;y<n;y++){
                if(a[y]==x){
                    sortpl=x;
                    cou++;
                }
                if(cou==pl-mi)
                    break;
            }
            if(cou==pl-mi)
                break;
        }
    }
    if(pl-mi<0){
        for(x=0;x>-1000000000;x--){
            for(y=0;y<n;y++){
                if(a[y]==x){
                    sortpl=x;
                    cou--;
                }
                if(cou==pl-mi)
                    break;
            }
            if(cou==pl-mi)
                break;
        }
    }

    if(pl-mi>0){
        printf("%d\n",an-(sortpl*n));
    }
    else if(pl-mi<0){
        printf("%d\n",an-(sortpl*n*-1));
    }
    else
        printf("%d\n",an);



    return 0;
}