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: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector< pair<int,int> > vpii;
typedef vector< pair<ll,ll> > vpll;
typedef vector< vector<int> > matrix;

#define SET(Ar,val) memset(Ar,val,sizeof(Ar))
#define rep(i,val,n)  for(int i=val;i<n;i++)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define LMAX LONG_LONG_MAX
#define LMIN LONG_LONG_MIN
#define IMAX INT_MAX
#define IMIN INT_MIN
#define M 1000000007ll
#define md 998244353

/*--------------------------------------------------------------------------------------------------------------------------------*/

template <class T> T modpower(T a,T b,T MOD)
{
    T ans=1;
    while(b!=0)
    {
        if(b%2==1)
        ans=(ans*a)%MOD;
        a=(a*a)%MOD;
        b/=2;
    }
    return ans;
}


void init()
{
    int N;
    ll x;
    vll v0;
    cin >> N;
//    int c0,cn,cp,_cp,_cn;
//    c0 = cn = cp = _cp = _cn = 0;
    rep(i,1,N+1)
    {
        cin >> x;
        x = x - i;
        v0.pb(x);
//        se1.insert(x);
//        if(x < 0)
//        {
//            cn += (x);
//            _cn++;
//        }
//        else if(x > 0)
//        {
//            cp += x;
//            _cp++;
//        }
//        else
//        {
//            c0++;
//        }
    }
    sort(v0.begin(),v0.end());
    if(N%2)
    {
        x = v0[N/2];
    }
    else
    {
        x = (v0[N/2] + v0[N/2 - 1])/2;
    }
    ll ans  = 0;
    for(int i=0;i<N;i++)
    {
        ans += abs(x-v0[i]);
    }
    cout << ans  ;
}


int main()
{
    clock_t time1 = clock();
//	#ifndef ONLINE_JUDGE
//	freopen("input1.txt","r",stdin);
//	#else
//	//ONLINE_JUDGE
//	#endif
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
    init();
    clock_t time2 = clock();
    cerr <<"Time: " <<(double)(time2-time1)/CLOCKS_PER_SEC * 100000 <<" seconds" <<endl;
	return 0;
    }