Some time ago Mister B detected a strange signal from the space, which he started to study.

After some transformation the signal turned out to be a permutation p of length n or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.

Let's define the deviation of a permutation p as <image>.

Find a cyclic shift of permutation p with minimum possible deviation. If there are multiple solutions, print any of them.

Let's denote id k (0 ≤ k < n) of a cyclic shift of permutation p as the number of right shifts needed to reach this shift, for example:

  * k = 0: shift p1, p2, ... pn, 
  * k = 1: shift pn, p1, ... pn - 1, 
  * ..., 
  * k = n - 1: shift p2, p3, ... pn, p1. 

Input

First line contains single integer n (2 ≤ n ≤ 106) — the length of the permutation.

The second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of the permutation. It is guaranteed that all elements are distinct.

Output

Print two integers: the minimum deviation of cyclic shifts of permutation p and the id of such shift. If there are multiple solutions, print any of them.

Examples

Input

3
1 2 3


Output

0 0


Input

3
2 3 1


Output

0 1


Input

3
3 2 1


Output

2 1

Note

In the first sample test the given permutation p is the identity permutation, that's why its deviation equals to 0, the shift id equals to 0 as well.

In the second sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 2, 3) equals to 0, the deviation of the 2-nd cyclic shift (3, 1, 2) equals to 4, the optimal is the 1-st cyclic shift.

In the third sample test the deviation of p equals to 4, the deviation of the 1-st cyclic shift (1, 3, 2) equals to 2, the deviation of the 2-nd cyclic shift (2, 1, 3) also equals to 2, so the optimal are both 1-st and 2-nd cyclic shifts.
import java.util.Scanner;

public class D {

    private static Scanner in;

    private static int n;
    private static long[] p, d1, d2, d3;

    private static void solve() {
        n = in.nextInt();
        p = new long[n + 1];
        d1 = new long[n];
        d2 = new long[n];
        d3 = new long[n];
        for (int i = 1; i <= n; i++) {
            p[i] = in.nextLong();
        }
        for (int i = 1; i <= n; i++) {
            if (i < p[i]) {
                d1[(int)(p[i] - i - 1)] += 2;
                d1[((n - i - 1) + n) % n] += (Math.abs(p[i] - 1) - Math.abs(n - p[i])) - 1;
                d1[n - i] += -1 - (Math.abs(p[i] - 1) - Math.abs(n - p[i]));
            } else {
                d1[((n - i - 1) + n) % n] += (Math.abs(p[i] - 1) - Math.abs(n - p[i])) - 1;
                d1[n - i] += -1 - (Math.abs(p[i] - 1) - Math.abs(n - p[i]));
                d1[(int) (p[i] - i + n - 1)] += 2;
            }
            d2[0] += Math.abs((i % n + 1) - p[i]) - Math.abs(i - p[i]);
        }
        for (int i = 1; i < n; i++)
            d2[i] = d2[i - 1] + d1[i - 1];
        for (int i = 1; i <= n; i++)
            d3[0] += Math.abs(p[i] - i);
        for (int i = 1; i < n; i++)
            d3[i] = d3[i - 1] + d2[i - 1];

        long min = d3[0];
        int minI = 0;
        for (int i = 0; i < n; i++) {
            if (d3[i] < min) {
                min = d3[i];
                minI = i;
            }
        }
        System.out.println(min + " " + minI);
    }

    public static void main(String[] args) {
        in = new Scanner(System.in);
        solve();
        System.exit(0);
    }

}
