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.*;
import java.util.*;

// Not finished
public class d820 {

	public static void main(String[] args) throws IOException {
		IO io = new IO(System.in);

		int n = io.nextInt();

		int[] p = new int[n];
		for (int i = 0; i < p.length; i++) {
			p[i] = io.nextInt() - 1;
		}

		int[] pm = new int[n+1];

		for (int i = 0; i < n; i++) {
			if (p[i] >= i) {
				pm[0]--;
				pm[p[i] - i] += 2;
				pm[n - i - 1] -= 1;
				pm[n - i]--;
			} else {
				pm[0]++;
				pm[n - i - 1]--;
				pm[n - i]--;
				pm[n - i + p[i]] += 2;
			}
		}

		long cur = 0;
		for (int i = 0; i < n; i++) {
			cur += Math.abs(p[i] - i);
		}

		long increasing = 0;
		long best = cur;
		long bestid = 0;
		for (int i = 0; i < n-1; i++) {
//			io.println(cur);
			increasing += pm[i];
			cur += increasing;
			cur = cur - (n - 1) + 2 * p[n - 1 - i]; // jump
			
			if (cur < best) {
				best = cur;
				bestid = i+1;
			}
		}
		
		io.printf("%d %d\n",best,bestid);

		io.close();
	}

	static class IO extends PrintWriter {
		static BufferedReader r;
		static StringTokenizer t;

		public IO(InputStream i) {
			super(new BufferedOutputStream(System.out));
			r = new BufferedReader(new InputStreamReader(i));
			t = new StringTokenizer("");
		}

		public String next() throws IOException {
			while (!t.hasMoreTokens()) {
				t = new StringTokenizer(r.readLine());
			}
			return t.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	}
}