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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <functional>
#include <queue>
#include <string>
#include <cstring>
#include <numeric>
#include <cstdlib>
#include <cmath>
using namespace std;

typedef long long ll;

#define INF 10e10
#define rep(i,n) for(int i=0; i<n; i++)
#define rep_r(i,n,m) for(int i=m; i<n; i++)
#define END cout << endl
#define MOD 1000000007
#define pb push_back
#define sorti(x) sort(x.begin(), x.end())
#define sortd(x) sort(x.begin(), x.end(), std::greater<int>())

 
ll a[200050] = {};
ll A[200050] = {};
 
int main()
{
	ll n;
	ll ans = 0;
	cin >> n;
  vector<ll> a(n), A(n);
	for (ll i = 0; i < n; i++)
	{
		cin >> a[i];
		A[i] = a[i];
	}
 
	sort(a.begin(), a.end());
	ll b;
	if (n % 2 != 0) {
		b = a[(n - 1) / 2];
	}
	else {
		b = (a[n / 2] + a[(n / 2) - 1]) / 2;
	}
 
	b = b - (n - 1) / 2;
 
	for (ll i = 0; i < n; i++)
	{
		ans += abs(A[i] - b);
		b++;
	}
 
	cout << ans << endl;
}