One day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:

a1, a2, ..., an → an, a1, a2, ..., an - 1.

Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

Input

The first line contains an integer n (2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

Examples

Input

2
2 1


Output

1


Input

3
1 3 2


Output

-1


Input

2
1 2


Output

0
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class LittlePonyShiftSort {

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());

        StringTokenizer st = new StringTokenizer(br.readLine());

        int a[] = new int[n];

        for(int i=0; i<n; i++){
            a[i] = Integer.parseInt(st.nextToken());
        }

        boolean firstFound = false;
        boolean secondFound = false;
        int index = -1;

        for(int i=0; i<n-1 && !secondFound; i++){
            if(a[i+1]<a[i]){
                if(firstFound)
                    secondFound = true;
                else{
                    firstFound = true;
                    index = i;
                }
            }
        }

        if(secondFound)
            System.out.println(-1);
        else
            if(firstFound){
                if(a[n-1]>a[0])
                    System.out.println(-1);
                else
                    System.out.println(a.length-index-1);
            }
            else
                System.out.println(0);

    }

}
