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>
#define REP(i, n) for(long long i=0; i<n; i++)
#define REPR(i, n) for(long long i=n-1; i>=0; i--)
#define FOR(i, m, n) for(long long i=m; i<=n; i++)
#define FORR(i, m, n) for(long long i=m; i>=n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.rbegin(), v.rend());
#define ALL(v) (v).begin(),(v).end()
#define FIN ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;
const ll mod = 1e9+7;
const ll inf = 1e15;


ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}


int main(){FIN

    ll n;cin>>n;
    vll a(n);
    REP(i,n) {
        cin >> a[i];
        a[i] -= i-1;
    }
    VSORT(a);
    ll b=0;
    if (a.size()%2==1) b = a[(n-1)/2];
    else b=a[n/2];
    ll ans =0;
    REP(i,n) {
        ans += abs(a[i]-b);
    }
    cout<<ans<<endl;
    return 0;
}