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: Accepted)

#include <stdio.h>
int n;
long a[500000];
int main(){
	scanf("%d",&n);
	long long sum=0;
	for (int i=1;i<=n;i++){
		scanf("%li",&a[i]);
		a[i]=a[i]-i;	
	}
	sort(1,n);
	a[0]=a[1];
	for (int i=1;i<=n;i++){
		a[i]=a[i]-a[0];
		sum=sum+a[i];
	}
	long long ans=sum;
	for (int i=2;i<=n;i++){
		sum=sum-(a[i]-a[i-1])*(n-i+1)+(a[i]-a[i-1])*(i-1);
		if (sum<ans)ans=sum;
	}
	printf("%lli",ans);
	return(0);
}
int sort(int x,int y){
	int l=x;
	int r=y;
	long m=a[(x+y)/2];
	while (l<r){
		while (a[l]<m)l++;
		while (a[r]>m)r--;
		if (l<=r){
			long z=a[l];
			a[l]=a[r];	
			a[r]=z;
			l++;
			r--;
		}
	}
	if (l<y)sort(l,y);
	if (r>x)sort(x,r);
}