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<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
	int N;
	cin >> N;
	int A[N];
	for(int i=0; i<N; i++){
		cin >> A[i];
		A[i] -= i+1;
	}
	if(N%2 == 0){
		ll a = 0, b = 0;
		for(int i=0; i<N; i++){
			a += abs(A[i] - A[(N/2) - 1]);
		}
		for(int i=0; i<N; i++){
			b += abs(A[i] - A[N/2]);
		}
		cout << min(a, b) << endl;
	}else{
		ll a = 0;
		for(int i=0; i<N; i++){
			a += abs(A[i] - A[N/2]);
		}
		cout << a << endl;
	}
	return 0;
}