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

public class File {
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		int n = sc.nextInt();
		int[] a = new int[n];
		
		boolean allEqual = true;
		
		int min = Integer.MAX_VALUE;
		int minIndex = -1;
		
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			
			if (i > 0 && a[i] != a[i-1]) {
				allEqual = false;
			}
			
			if (a[i] < min) {
				min = a[i];
				minIndex = i;
			}
		}
		
		if (allEqual) {
			out.println(0);
		}
		else if (n == 2) {
			int res = 0;
			if (a[0] > a[1]) {
				res++;
			}
			out.println(res);
		}
		else {
			boolean isPossible = true;
			
			// Find the leftmost index of the min value (in cycle).
			int index = minIndex;			
			
			while (a[index] == min) {
				minIndex = index;
				index = index == 0 ? n-1 : index-1;
			}
			
			index = minIndex;
			
			for (int i = 0; i + 1 < n; i++) {
				int val = a[index];
				int right = index == n-1 ? a[0] : a[index+1];
				
				if (val > right) {
					isPossible = false;
					break;
				}
				
				index = (index + 1) % n;
			}
			
			
			if (isPossible) {
				out.println((n - minIndex) % n);
			}
			else {
				out.println(-1);
			}
		}
		
		
		
		
		out.close();
	}	
}






















