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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define fst first
#define snd second
#define SZ(u) ((int) (u).size())
#define ALL(u) (u).begin(), (u).end()

inline void proc_status()
{
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline T read()
{
	register T sum(0), fg(1);
	register char ch(getchar());
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') fg = -1;
	for(;  isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) - '0' + ch;
	return sum * fg;
}

typedef long long LL;
typedef pair<int, int> pii;

const int MAXN = (int) 2e5;

int n;

int a[MAXN + 5];

inline void input()
{
	n = read<int>();
	for(int i = 1; i <= n; ++i) a[i] = read<int>() - i;
}

inline void solve()
{
	sort(a + 1, a + n + 1);

	int mid = (n + 1) / 2;
	LL ans = 0;
	for(int i = 1; i <= n; ++i) ans += abs(a[i] - a[mid]);
	printf("%lld\n", ans);
}

int main()
{
//#ifndef ONLINE_JUDGE
//	freopen("C.in", "r", stdin);
//	freopen("C.out", "w", stdout);
//#endif

	input();
	solve();

	return 0;
}