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)

#include <bits/stdc++.h>
using namespace std;
struct Fast {
    Fast() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);
    }
} fast;
template <typename A, size_t N, typename T> void Fill(A (&array)[N], const T& val) {
    std::fill((T*)array, (T*)(array + N), val);
}
// long long mod = 998244353;
#define REP(i, n) for (long long i = 0; i < (n); i++)
#define rep(i, a, n) for (long long i = a; i < (n); i++)
typedef long long ll;
typedef pair<ll, ll> P;
long long mod = 1000000007;
long long modpow(long long m, long long n, long long modvalue) {
    if (n == 0) return 1;
    if (n % 2 == 0) {
        long long t = modpow(m, n / 2, modvalue);
        return t * t % modvalue;
    } else {
        return modpow(m, n - 1, modvalue) * m % modvalue;
    }
}

long long moddiv(long long a, long long b, ll modvalue) { return (a * modpow(b, mod - 2, modvalue)) % modvalue; }

const ll INF = 1LL << 60;

string change_binary(long long a) {
    string ret;
    for (int i = 31; i >= 0; i--) {
        ret += (a >> i & 1) + '0';
    }
    return ret;
}

ll nCkmod(ll n, ll k, ll modvalue) {
    ll numera = 1;
    for (int i = n; i > n - k; i--) {
        numera *= i;
        numera %= modvalue;
    }
    ll denomi = 1;
    REP(i, k) {
        denomi *= (i + 1);
        denomi %= modvalue;
    }
    return moddiv(numera, denomi, modvalue);
}

map<ll, ll> cnt;
void primeFactors(long long n) {
    while (n % 2 == 0) {
        cnt[2]++;
        n = n / 2;
    }
    for (int i = 3; i <= sqrt(n); i = i + 2) {
        while (n % i == 0) {
            cnt[i]++;
            n = n / i;
        }
    }
    if (n > 2) {
        cnt[n]++;
    }
}
int main() {
    ll n;
    cin >> n;
    vector<ll> as(n);
    REP(i, n) {
        cin >> as[i];
        as[i] -= i + 1;
    }
    sort(as.begin(), as.end());
    ll sum = 0;
    REP(i, n) { sum += abs(as[i] - as[n / 2]); }
    cout << sum << endl;
}