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.io.PrintWriter;
import java.util.*;

public class B_454{
    static PrintWriter pw = new PrintWriter(System.out);

    public static void main(String arg[]) throws IOException {
        InputStreamReader isr = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);
        StringTokenizer token = new StringTokenizer(" ");


        int n=Integer.parseInt(br.readLine());
        int i=0;
        int arr[]=new int[n];

        token=new StringTokenizer(br.readLine());
        for(i=0;i<=n-1;i++)
            arr[i]=Integer.parseInt(token.nextToken());



        pw.println(getResult(n,arr));
        pw.close();

    }

    private static long getResult(int n,int arr[]) {

        int i=0;

        while (i<=n-2 && arr[i]<=arr[i+1])
            i++;

        if(i==n-1)
            return 0;

        int p=i;
        int q=i+1;
        i=i+1;

        while (i<=n-2 && arr[i]<=arr[i+1])
            i++;

        if(i==n-1){
            if(arr[n-1]<=arr[0])
                return n-q;
            else
                return -1;
        }
        else{
            //arr[i]>arr[i+1]
            return -1;
        }
    }


}