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;

/**
 * Created by shalin on 2/7/17.
 */
public class Problem819B {

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int[] p = new int[n];
    for (int i = 0; i < n; i++) {
      p[i] = in.nextInt();
    }

    long[] d = new long[n + 1];
    long[] df = new long[n + 1];

    for (int k = 0; k < n; k++) {
      if (p[k] <= k) {
        add(d, df, 0, n - (k + 1), 1, k + 1 - p[k]);
        add(d, df, n - k, n - (k + 1) + p[k], -1, p[k] - 1);
        add(d, df, n - (k + 1) + p[k] + 1, n - 1, 1, 1);
      } else {
        add(d, df, 0, p[k] - (k + 1), -1, p[k] - (k + 1));
        add(d, df, p[k] - k, n - (k + 1), 1, 1);
        add(d, df, n - k, n - 1, -1, p[k] - 1);
      }
    }

    long[] res = new long[n];
    long sumdf = 0;
    for (int i = 0; i < n; i++) {
      d[i] += sumdf;
      sumdf += df[i];
    }

    long sum = 0;
    for (int i = 0; i < n; i++) {
      sum += d[i];
      res[i] += sum;
    }

    int minI = 0;
    for (int i = 0; i < n; i++) {
      if (res[minI] > res[i]) {
        minI = i;
      }
    }

    System.out.println(res[minI] + " " + minI);
  }

  static void add(long[] d, long[] df, int l, int r, int cd, int v) {
    if (l > r) return;

    d[l] += v;
    d[r + 1] -= v + cd * 1L * (r - l);
    df[l] += cd;
    df[r] -= cd;
  }
}
