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 Div2_421D {

	static final int INCR = 1;
	static final int DECR = -1;

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		int numI = Integer.parseInt(reader.readLine());
		int[] items = new int[numI + 1];

		int[] numFurth = new int[2 * (numI + 1)];

		long delta = 0;

		int cNumC = 0;
		int cNumF = 0;

		StringTokenizer inputData = new StringTokenizer(reader.readLine());
		for (int i = 1; i <= numI; i++) {
			items[i] = Integer.parseInt(inputData.nextToken());
			delta += Math.abs(items[i] - i);

			if (items[i] > i) {
				cNumC++;
				numFurth[items[i] - i + 1]++;
			} else {
				cNumF++;
			}
		}

		long ans = delta;
		int ansS = 0;
		for (int cRot = 1; cRot < numI; cRot++) {
			cNumC -= numFurth[cRot];
			cNumF += numFurth[cRot];
			delta += cNumF - cNumC;

			// items[numI - cRot] is a special case
			delta -= Math.abs(items[numI - cRot + 1] - (numI + 1));
			delta += items[numI - cRot + 1] - 1;

			if (items[numI - cRot + 1] > 1) {
				cNumF--;
				cNumC++;
				numFurth[cRot + items[numI - cRot + 1]]++;
			}
			if (delta < ans) {
				ans = delta;
				ansS = cRot;
			}
		}
		System.out.println(ans + " " + ansS);
	}

}
