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.Scanner;

public class D {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner in = new Scanner(System.in);
		
		int n=in.nextInt();
		
		int[] p=new int[n+1];
		
		for (int i=1;i<=n;i++)
			p[i]=in.nextInt();
		
		long[] diff= new long[n];
		long temp=0;
		long res=0;
		for (int i=1;i<=n;i++){
			res+=Math.abs(p[i]-i);
			if (p[i]<=i){
				temp++;
				diff[n-i]-=2;
				if (p[i]!=i)
					diff[n-i+p[i]-1]+=2;
			}
			else{
				temp--;
				diff[p[i]-i-1]+=2;
				diff[n-i]-=2;
			}
		}
		
		long[] dif = new long[n];
		long[] dif2 = new long[n];
		dif[0]=temp;
		dif2[0]--;
		dif2[0]-=Math.abs(p[n]-n);
		dif2[0]+=Math.abs(p[n]-1);
		for (int i=1;i<n;i++){
			dif[i]=dif[i-1]+diff[i-1];
			dif2[i]--;
			dif2[i]-=Math.abs(p[n-i]-n);
			dif2[i]+=Math.abs(p[n-i]-1);
		}
		
		long minr=res;
		int mins=0;
		long[] val = new long[n];
		
		for (int i=0;i<n;i++){
			val[i]=res;
			res+=dif[i] + dif2[i];
//			res--;
//			res-=Math.abs(p[n-i]-n);
//			res+=Math.abs(p[n-i]-1);
		}
		for (int i=0;i<n;i++){
			if (minr>val[i]){
				minr=val[i];
				mins=i;
			}			
//			System.out.printf("shift=%d    value=%d		dif=%d		dif2=%d		diff=%d\n", i,val[i],dif[i],dif2[i],diff[i]);
		}
		
		System.out.println(minr+" "+mins);
	}

}
