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.Arrays;
import java.util.Scanner;

public class R421D {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] arr = new int[n];
		int[] count = new int[n];
		int S = 0;
		long sum = 0;
		for(int i = 0; i < n; i++) {
			arr[i] = scan.nextInt()-1;
			sum += Math.abs(arr[i]-i);
			if(arr[i] >= i){
				int x = arr[i]-i;
				count[x]++;
			}
			else{
				int x = n-i+arr[i];
				count[x]++;
			}
			if(arr[i] <= i) S++; 
		}
		int idx = 0;
		long minSum = sum;
		for(int i = 1; i < n; i++) {
			sum -= Math.abs(n-arr[n-i]-1);
			sum += arr[n-i];
			sum += S-1;
			sum -= n-S;
			S--;
			S += count[i];
			//System.out.println(i+" "+sum);
			if(sum < minSum) {
				minSum = sum;
				idx = i;
			}
		}
		System.out.println(minSum+" "+idx);
	}
}

/*
6
4 3 1 2 5 6
*/