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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author HARSHAD
 */

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const ld PI = 3.14159265358979323846;

#define f(i,n) for(int i=0;i<n;i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)

template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

class A {
public:
	void solve(std::istream& in, std::ostream& out) {
		ll n; in>>n;
		ll arr[n], avg = 0;
		f(i, n) in>>arr[i], arr[i] -= (i+1), avg += arr[i];
		avg /= n;
		vl b;
		b = {avg-1, avg, avg+1};
		ll res = LLONG_MAX;
		for(int x: b){
			ll ans = 0;
			f(i, n) ans += abs(arr[i]-x);
			res = min(res, ans);
		}
		out<<res;
	}
};


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	A solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}