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<vector>
#include<algorithm>
#include<map>
#include<cstring>
#include<math.h>
#include<bitset>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
#include<assert.h>
#include<string>
#include<cstdlib>
#include<cctype>
#include<unordered_map>
#include<sstream>
// #include<bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int,int>;
constexpr ll INF = 1LL << 40;
const double INFD = 1e100;
const ll mod = 1000000007;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -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やllをstringに変換
template<typename T> inline string toString(const T &a){
  ostringstream oss;
  oss << a;
  return oss.str();
};
// ----------------------------------------------------------------------------

int main(){
  int N;
  cin >> N;
  vector<ll> A(N);
  for(int i=0; i<N; i++){
    cin >> A[i];
    A[i] -= i+1;
  }
  sort(A.begin(),A.end());
  ll ans = INF;
  if(N%2){
    ll mid = A[N/2];
    ans = 0;
    for(int i=0; i<N; i++){
      ans += abs(mid-A[i]);
    }
  }else{
    ll mid = (A[N/2]+A[N/2-1])/2;
    ll temp = 0;
    for(int i=0; i<N; i++){
      temp += abs(mid - A[i]);
    }
    chmin(ans,temp);
    mid++;
    temp = 0;
    for(int i=0; i<N; i++){
      temp += abs(mid - A[i]);
    }
    chmin(ans,temp);
  }
  cout << ans << endl;
  return 0;
}