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)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <numeric>
#include <math.h>
#include <algorithm>
#include <functional>
#include <string>
#include <array>
#include <vector>
#include <list>
#include <stack>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <deque>
#include <queue>
#include <complex>
#include <bitset>
#define full(c) c.begin(), c.end()
#define vector2d(name,type,h,w,init) vector<vector<type>> name(h,vector<type>(w,init)) 
const int PRIME = 1000000007;
const int INT_INF = 2147483647;
const long long int LL_INF = 9223372036854775807;
const double PI = acos(-1);
const double EPS = 0.000000001;
typedef unsigned long long int ull;
typedef long long int ll;
using namespace std;

int main(){
	//ios_base::sync_with_stdio(false);
	//cin.tie(NULL);
	int i, n, b, sum = 0;
	cin >> n;
	vector<int> a(n);
	for (i = 0; i < n; i++){
		cin >> a[i];
		a[i] -= i + 1;
	}
	sort(full(a));
	if (n % 2 == 0) b = (a[n / 2] + a[n / 2 - 1]) / 2;
	else b = a[n / 2];
	for (i = 0; i < n; i++) sum += abs(a[i] - b);
	cout << sum << endl;
	return 0;
}