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++ (GCC 9.2.1) (judge verdict: Accepted)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>

using namespace std;

//----------------------- Print Function ----------------------//

inline void print() {
    cout << endl;
}
template <typename First, typename... Rest>
void print(const First &first, const Rest &... rest) {
    cout << first << ' ';
    print(rest...);
}

//------------------------- Libraries -------------------------//

//--------------------------- Solve ---------------------------//

void solve() {
    int N; cin >> N;
    vector<long long> A(N);
    for (int i = 0; i < N; i++) {
        int a; cin >> a;
        A[i] = a - i - 1;
    }

    sort(A.begin(), A.end());

    vector<long long> cumsum(N+1);
    for (int i = 0; i < N; i++) {
        cumsum[i+1] += cumsum[i] + A[i];
    }

    long long ans = 1LL<<60;
    for (int i = 0; i <= N; i++) {
        long long tmp = -cumsum[i] + cumsum[N] - cumsum[i];

        long long add = 1LL<<60;
        if (i > 0) add = A[i-1] * (i - (N-i));
        if (i < N) add = min(add, A[i] * (i - (N-i)));
        tmp += add;

        ans = min(ans, tmp);
    }

    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    solve();

    return 0;
}