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 <cstdlib>
#include <algorithm>
using namespace std;

int main() {
	int N;
	cin >> N;
	int A[200001];
	int sumv = 0;
	for (int i = 0; i < N; i++) { 
		cin >> A[i];
		A[i] -= (i + 1);
		sumv += A[i];
	}
	int avev = sumv / N;
	int avev2 = avev + 1;
	int avev3 = avev -1;
	int avev4 = avev + 2;
	int avev5 = avev - 2;
	int avev6 = avev + 3;
	int avev7 = avev - 3;
	int ans1 = 0;
	int ans2 = 0;
	int ans3 = 0;
	int ans4 = 0;
	int ans5 = 0;
	int ans6 = 0;
	int ans7 = 0;
	for (int i = 0; i < N; i++) {
		ans1 += abs(A[i] - avev);
		ans2 += abs(A[i] - avev2);
		ans3 += abs(A[i] - avev3);
		ans4 += abs(A[i] - avev4);
		ans5 += abs(A[i] - avev5);
		ans6 += abs(A[i] - avev6);
		ans7 += abs(A[i] - avev7);
	}
	cout << min({ ans1, ans2, ans3, ans4, ans5, ans6, ans7 }) << endl;
}