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<math.h>
#include<string.h>
#include <time.h>
typedef long long llint;
typedef long double ldouble;
// 小大
int compare(const void *a, const void *b){return *(llint *)a - *(llint *)b;}
typedef struct{
    llint aa;
    llint bb;
}frequent;
// 小大
int cmp( const void *p, const void *q ) {
    return ((frequent*)p)->bb - ((frequent*)q)->bb;
}
llint ngauss(llint a,llint b){
    // printf("%lld %lld",a-a/b*a,(a/b+1)*a-a);
    return (a-a/b*a>(a/b+1)*b-a)?a/b+1:a/b;
}
llint antingauss(llint a,llint b){
    // printf("%lld %lld",a-a/b*a,(a/b+1)*a-a);
    return (a-a/b*a>(a/b+1)*b-a)?a/b:a/b+1;
}
/*絶対値*/
llint ABS(llint a){return a>=0?a:(-a);}



int main(void){
    llint n,i,ave=0,ans=0,div,aa=0,j;
    scanf("%lld",&n);
    llint a[n];
    for(i=0;i<n;i++){
        scanf("%lld",&a[i]);
        a[i]-=i+1;
        ave+=a[i];
    }
    // printf("%lld ",ave);
    div = ave/n;
    // printf("%lld ",div);
    div-=4;
    for(j=0;j<9;j++){
        div++;
        aa=0;
        for(i=0;i<n;i++){
            aa+=ABS(a[i]-div);
        }
        // printf("%lld\n",aa);
        if(j==0) ans=aa;
        if(aa<ans) ans=aa;
    }
    // for(i=0;i<n;i++){
    //     // printf("%lld ",a[i]);
    //     // printf("%lld\n",ABS(a[i]-i-1-div));
    //     ans+=ABS(a[i]-div);
    // }
    // div = antingauss(ave,n);
    // printf("%lld ",div);
    // for(i=0;i<n;i++){
    //     // printf("%lld ",a[i]);
    //     // printf("%lld\n",ABS(a[i]-i-1-div));
    //     aa+=ABS(a[i]-div);
    // }
    // if(aa<ans) ans=aa;



    printf("%lld\n",ans);
    return 0;
}