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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using std::sort;
using std::min;
#define N 203010
#define ll long long
int a[N];
ll abs(ll a) {return a>0?a:-a;}
int main()
{
	int n;ll sum=0,ans1=0,ans2=0,B1,B2;
	scanf("%d",&n);
	for (int i=1;i<=n;++i) scanf("%d",&a[i]),a[i]-=i;
	sort(a+1,a+1+n);
	if (n&1) B1=B2=a[(n>>1)+1];else B1=floor((a[n>>1]+a[(n>>1)+1])/2.0),B2=ceil((a[n>>1]+a[(n>>1)+1])/2.0);
//	B1=floor(1.0*sum/n);B2=ceil(1.0*sum/n);
//	printf("%lf %d %d\n",sum/n,B1,B2);
	for (int i=1;i<=n;++i) ans1+=abs(a[i]-B1),ans2+=abs(a[i]-B2);
	printf("%lld",min(ans1,ans2));
	return 0;
}