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.*;

public class Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int n = sc.nextInt();
              int arr[] = new int[n];
              int pos[] = new int[n+1];
              for(int i = 0 ; i < n ; i++)
              {
                     arr[i] = sc.nextInt();
                     pos[arr[i]] = i;
              }
              
              int diff[] = new int[n];
              for(int i = 1 ; i <= n ; i++)
              {
                     diff[(i-1-pos[i]+n)%n]++;
              }
              long left = 0 , right = 0;
              long dev = 0;
              long min = Long.MAX_VALUE;
              
             for(int i = 1 ; i <= n ; i++)
             {
                    dev += Math.abs((pos[i]) - (i-1));
                    if(pos[i] < i-1)
                    left++;
                    
                    else
                    right++;
             }
             
             min = Math.min(min , dev);
             int id = 0;
             
              
             for(int i = 1 ; i < n ; i++)
             {
                   
                 
                 dev += Math.max(0 , right-1);
                 dev -= left;
                 long x = -(n-arr[n-i]) + arr[n-i]-1; 
                 dev = dev +x;
                 if(dev < min)
                 {
                        min = dev;
                        id = i;
                 }
                 if((pos[1] + i)%n == 0)
                 right += Math.max(diff[i]-1 , 0);
                 
                 else
                 right += diff[i];
                 
                 if(arr[n-i] != 1)
                 right--;
                 
                 left = n-right;
              
             }
            
             System.out.println(min + " " + id);
       }
}