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;
using ll=long long;
const int INF = 100000000;
ll i,j,k;
ll mod=1000000007;

int main(){
    ll n,answer=0,answer2=0;cin>>n;
    vector<ll> a(n+1),key(n+1);
    
    for(i=1;i<=n;i++){cin>>a[i];}
    for(i=1;i<=n;i++){key[i]=a[i]-i;}

    double average=0;
    ll llaverage,llaverage2;

    for(i=1;i<=n;i++){
        average+=(key[i]+.0);
    }
    average/=(n+.0);


    if(average<0){
        llaverage=(ll)(average)-1;
        llaverage2=(ll)(average);
    }else{
        llaverage=(ll)(average);
        llaverage2=(ll)(average)+1;
    }


    for(i=1;i<=n;i++){
        answer+=abs(key[i]-llaverage);       
        answer2+=abs(key[i]-llaverage2);
    }
    
    cout<<min(answer,answer2)<<endl;
    return 0;
}