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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define REP(i, n)  FOR(i, 0, n)
#define ASC(c) sort((c).begin(), (c).end())
#define DESC(c) sort((c).begin(), (c).end(), greater<int>())

#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define DEBUG(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

typedef long long ll;
typedef unsigned long long ull;
struct edge { int u, v; ll w; };
 
ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-10;

int main() {
  int n;cin >> n;
  int  a[2 * 100000 + 1];
  for(int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }

  sort(a + 1, a + n + 1);
  int b1 = a[n / 2 + 1];
  int b2 = a[n / 2];
  int ans1 = 0;
  int ans2 = 0;
  for(int i = 1; i <= n; i++) {
    ans1 += abs(a[i] - b1);
    ans2 += abs(a[i] - b2);
  }

  cout << min(ans1, ans2) << endl;

}