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;

#define INF 1001001001
#define LINF 1001001001001001001
#define MOD 1000000007
#define MOD2 998244353

template<class T,class U>bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U>bool chmin(T &a, const U &b){if(b<a){a=b;return 1;}return 0;}

#define rep(i,n) for(int i=0,_i=(n);i<_i;++i)
#define rep1(i,a,b) for(int a_=(a),b_=(b),i=a_;i<b_;++i)
#define repr(i,n) for(int _i=(n),i=_i;i>0;--i)

#define db(x) cerr<<#x<<" = "<<x<<" ";
#define db2(x,y) cerr<<"("<<#x<<", "<<#y<<") = ("<<x<<", "<<y<<") ";
#define db3(x,y,z) cerr<<"("<<#x<<", "<<#y<<", "<<#z<<") = ("<<x<<", "<<y<<", "<<z<<") ";
#define ln cout<<endl;

#define all(a)  (a).begin(),(a).end()
#define dig(n)  to_string(n).length()

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define se second
#define fi first

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pii,pii> ppii;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return (a*b)/gcd(a,b);}

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main(){
    bool flag=false;
    ll ans=0,sum=0;

    int n;
    cin >>n;
    vector<int>a(n);

    rep(i,n){
        cin>>a[i];
        a[i] -= i+1;
        sum += a[i];
    }

    sort(all(a));

    ll div = (ll)n*a[n/2];

    //cout <<fixed<<setprecision(16)<< << endl;
    cout<<sum - div<<endl;
    //if(flag)cout << "Yes" <<endl;
    //else cout << "No" <<endl;
    
    return 0;
}