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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Codeforces820D {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		int[] perm = new int[n];
		for (int i = 0; i < n; i++) {
			perm[i] = Integer.parseInt(st.nextToken());
		}
		
		long[] secondDiff = new long[n];
		for (int i = 0; i < n; i++) {
			if (perm[i] == 1) {
				secondDiff[n-i-1] += (long) n;
				secondDiff[(n-i)%n] -= (long) n;
			}
			else if (perm[i] == n) {
				secondDiff[n-i-1] -= (long) n;
				secondDiff[(n-i)%n] += (long) n;
			}
			else {
				int r = perm[i];
				secondDiff[n-i-1] += (long) (n-2*r+2);
				secondDiff[(n-i)%n] += (long) (2*r-n);
				secondDiff[(n+r-i-1)%n] -= (long) 2;
			}
		}
		
		long[] actval = new long[n];
		for (int i = 0; i < n; i++) {
			actval[0] += (long) Math.abs(i+1-perm[i]);
			actval[1] += (long) Math.abs(i+1-perm[(n+i-1)%n]);
		}
		for (int i = 2; i < n; i++) {
			actval[i] = (long) (2*actval[i-1]-actval[i-2]-secondDiff[i-1]);
		}
		
		int minI = 0;
		long minJ = actval[0];
		for (int i = 1; i < n; i++) {
			if (actval[i] < minJ) {
				minI = i;
				minJ = actval[i];
			}
		}
		
		System.out.println(minJ + " " + minI);

	}

}
