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<vector>
#include<math.h>
#include<queue>

#define MOD 1000000007
#define rep(i, n) for(ll i=0; i < (n); i++)

using namespace std;
typedef long long int ll;
typedef vector<ll> llvec;
typedef vector<double> dvec;
typedef pair<int, int> P;
typedef priority_queue<ll, vector<ll>, greater<ll> > gpq_int;
typedef priority_queue<ll, vector<ll>, less<ll> > lpq_int;


ll mod(ll a, ll mod){
  ll res = a%mod;
  if(res<0)res=res + mod;
  return res;
}

ll modpow(ll a, ll n, ll mod){
  ll res=1;
  while(n>0){
    if(n&1) res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll modinv(ll a, ll mod){
  return modpow(a, mod-2, mod);
}

ll N;
ll A[1000000];

int main(void){
  cin >> N;
  ll ans=1e16;
  rep(i,N){
    cin >>A[i];
    A[i]-=(i+1);
  }
  ll ave;
  rep(i, N){
    ave+=A[i];
  }
  ave = (int)ave/N;

  rep(ii, 3){
    ll rescandidate=0;
    int i = -1 + ii;
    ll tmp=ave + i;
    rep(j, N){
      rescandidate += abs(tmp - A[j]);
    }
    if(rescandidate<ans) ans = rescandidate;
  }
  cout << ans;
  
  return 0;
}