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.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class D {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();

        long[] diff = new long[n + 1];
        int[] inc = new int[n + 1];

        for (int i = 1; i <= n; ++i) {
            int p = scan.nextInt();

            if (i <= p) {
                diff[0] += (p - i);
                inc[0] -= 1;
                inc[p - i] += 2;

                inc[n - i] -= 1;

                diff[n - i + 1] -= (n - p);
                diff[n - i + 1] += (p - 1);
                inc[n - i + 1] -= 1;

            } else {
                // i > p
                diff[0] += (i - p);
                inc[0] += 1;

                diff[n - i + 1] -= (n - p);
                diff[n - i + 1] += (p - 1);
                inc[n - i] -= 1;
                inc[n - i + 1] -= 1;

                inc[n - (i - p)] += 2;

            }

            //System.out.println(str(diff));
            //System.out.println(str(inc));
        }

        long min = Long.MAX_VALUE;
        int bestShift = 0;

        long level = 0;
        int slope = 0;
        for (int i = 0; i < n; ++i) {
            level += diff[i];
            if (level < min) {
                min = level;
                bestShift = i;
            }
            slope += inc[i];
            level += slope;
        }

        System.out.println(min + " " + bestShift);
    }

    public static String str(int[] arr) {
        StringBuilder result = new StringBuilder();
        result.append("[");
        boolean first = true;
        for (int x: arr) {
            if (!first)
                result.append(",");

            result.append(x);
            first = false;
        }
        result.append("]");
        return result.toString();
    }
}
