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: Time Limit Exceeded)

#include<stdio.h>
int main()
{
	long long int a[200001],n,i,j,b,min,max,ans;
	scanf("%lld",&n);
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		a[i] -= i;
	}
	int m = (n%2==0?n/2:(n-1)/2);
	for(i=1;i<=m;i++){
		for(j=i+1;j<=n-i;j++){
			if(a[j]<a[i]){
				min = a[j];
				a[j] = a[i];
				a[i] = min;
			}
			if(a[j]>a[n-i+1]){
				max = a[j];
				a[j] = a[n-i+1];
				a[n-i+1] = max;
			}
		}
	}
	if(n%2==1){
		b = a[(n+1)/2];
	}if(n%2==0 && (a[n/2+1]-a[n/2])%2==1){
		b = (a[n/2+1]+a[n/2]+1)/2;
	}if(n%2==0 && (a[n/2+1]-a[n/2])%2==0){
		b = (a[n/2+1]+a[n/2])/2;
	}
	for(i=1;i<=n;i++){
		ans += abs(a[i]-b);
	}
	printf("%ld",ans);
	return 0;
}