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)

int N,A[2<<17];
void sort_impl(int l,int r)
{
	if(r-l<=1);
	else if(r-l==2)
	{
		if(A[l+1]<A[l])
		{
			int tmp=A[l];
			A[l]=A[l+1];
			A[l+1]=tmp;
		}
	}
	else
	{
		int a=A[l];
		int b=A[(l+r)/2];
		int c=A[r-1];
		if((b<a)!=(b<c))
		{
			A[l]=b;
			A[(l+r)/2]=a;
		}
		else if((c<a)!=(c<b))
		{
			A[l]=c;
			A[r-1]=a;
		}
		int pivot=A[l];
		int mid=l;
		for(int i=l+1;i<r;++i)
		{
			if(A[i]<pivot)
			{
				mid++;
				if(i!=mid)
				{
					int tmp=A[mid];
					A[mid]=A[i];
					A[i]=tmp;
				}
			}
		}
		if(mid>l)
		{
			int tmp=A[l];
			A[l]=A[mid];
			A[mid]=tmp;
		}
		sort_impl(l,mid);
		sort_impl(mid+1,r);
	}
}
void sort(int N)
{
	sort_impl(0,N);
}
main()
{
	scanf("%d",&N);
	for(int i=0;i<N;i++)scanf("%d",&A[i]),A[i]-=i;
	sort(N);
	long ans=0,mid=A[N/2];
	for(int i=0;i<N;i++)ans+=A[i]<mid?mid-A[i]:A[i]-mid;
	printf("%lld\n",ans);
}