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 <algorithm>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <iomanip>
#include <set>
using namespace std;

#define rep(i, s) for (int i = 0; i < s; ++i)
#define all(v) (v.begin(), v.end())
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define VEC(a, n) vector<int>a(n)
#define PQ(a) priority_queue<int>a
#define PQmin(a) priority_queue< int, :vector<int>, greater<int> >a
#define PAIR pair<int, int>
/*
sort(sp.begin(), sp.end(), [](PAIR l, PAIR r){
        return l.first<r.first || (l.first==r.first && l.second > r.second);
        });
*/

typedef long long  ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

int main() {
    int n;
    cin>>n;
    vector<int>a(n);
    for(int i=0;i<n;i++){
        int x;
        cin>>x;
        a.at(i)=x-(i+1);
    }
    double AVE=0;
    for(int i=0;i<n;i++){
        AVE+=a.at(i);
    }
    AVE/=n;
    int ans=1000000000;
    for(int i=-1000;i<=1000;i++){
        int SUM=0;
        for(int j=0;j<n;j++){
            SUM+=abs(a.at(j)-(int)AVE+i);
        }
        ans=min(ans,SUM);
    }
    cout<<ans<<endl;





    return 0;
}