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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fcout cout << fixed << setprecision(10)
#define rep(i,n)    for(int i=0; i<(int)n; i++)
#define mp(a,b) make_pair(a,b)
#define pb push_back
const ll inf = 1e18;
const ll mod = 1e9 + 7;

ll pow (ll x, ll n){
    ll ans = 1;
    while(n>0){
        if(n & 1 == 1){
            ans *= x;
        }
        x = x * x;
        n >>= 1;
    }
    return ans;
}


int main(){
    int n;   cin >> n;
    vint a(n);   rep(i,n)   cin >> a[i];
    rep(i,n){
        a[i] = a[i] - (i + 1);
    }

    int ave =0;
    rep(i,n){
        ave += a[i];
    }
    ave /= n;

    ll ans = inf;
    for(int b = ave - 10000; b < ave+10000; b++){
        ll c=0;
        rep(j,n){
            ll d = a[j] - b;
            c += abs(d);
        }
        ans = min(ans, c);
    }
    cout << ans << endl;
}