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: Runtime Error)

#include <bits/stdc++.h>
 
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define CLR(a) memset((a), 0 ,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vii;
typedef vector<string> vs;
typedef pair<int, int> pi;
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MKP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
const int INF = 999999;
const double EPS = 1e-10;
const double PI = acos(-1.0);
const ll MOD = 1000000007;
 
 
 
ll N,sum;
vll A, ans;
 
int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> N;
	REP(i, N) {
		ll x;
        cin >> x;
		A.push_back(x);
	}
	REP(i, N) {
		sum = 0;
		REP(j, std::accumulate(A.begin(), A.end(), 0) / A.size()) {
			sum += abs(A[j] - j - 1);
		}
		ans.push_back(sum);
	}
	
	std::cout << *std::min_element(ans.begin(), ans.end()) << endl;
	
	return 0;
}