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<string>
#include<iostream>
#include<algorithm>
#include<set>
#include<math.h>
#include<vector>
#include<sstream>
#include<queue>
#include<functional>
#include<bitset>
#include<cstdio>
#include<iomanip>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include <string.h>
using ll = long long;
 
#define rep(i, n) for(ll i = 0; i < (int)(n); i++)
#define reps(i,x) for(ll i=1;i<=(int)(x);i++)
#define rrep(i,x) for(ll i=((int)(x)-1);i>=0;i--)
#define rreps(i,x) for(ll i=(int)(x);i>0;i--)
#define all(x) (x).begin(),(x).end()
#define m0(x) memset(x,0,sizeof(x))
#define vll vector<ll>
#define vi vector<int>
#define mp make_pair
#define vpll vector<pair<ll,ll>>
#define vpi vector<pair<int,int>>
#define mod 1000000007 
using namespace std;
int a[200000];
int main() {
  ll sum = 0, n;
  cin >> n;
  rep(i, n){
    cin >> a[i];
    a[i] = a[i] -i -1;
  }
  sort(a, a+n);
  int id=round(n*1.0/2)-1;
  int md = a[id];
  rep(i,n)
    sum += abs(a[i]-md);
  cout <<sum<< endl;
}