Polycarp likes arithmetic progressions. A sequence [a_1, a_2, ..., a_n] is called an arithmetic progression if for each i (1 ≤ i < n) the value a_{i+1} - a_i is the same. For example, the sequences [42], [5, 5, 5], [2, 11, 20, 29] and [3, 2, 1, 0] are arithmetic progressions, but [1, 0, 1], [1, 3, 9] and [2, 3, 1] are not.

It follows from the definition that any sequence of length one or two is an arithmetic progression.

Polycarp found some sequence of positive integers [b_1, b_2, ..., b_n]. He agrees to change each element by at most one. In the other words, for each element there are exactly three options: an element can be decreased by 1, an element can be increased by 1, an element can be left unchanged.

Determine a minimum possible number of elements in b which can be changed (by exactly one), so that the sequence b becomes an arithmetic progression, or report that it is impossible.

It is possible that the resulting sequence contains element equals 0.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the number of elements in b.

The second line contains a sequence b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^{9}).

Output

If it is impossible to make an arithmetic progression with described operations, print -1. In the other case, print non-negative integer — the minimum number of elements to change to make the given sequence becomes an arithmetic progression. The only allowed operation is to add/to subtract one from an element (can't use operation twice to the same position).

Examples

Input

4
24 21 14 10


Output

3


Input

2
500 500


Output

0


Input

3
14 5 1


Output

-1


Input

5
1 3 6 9 12


Output

1

Note

In the first example Polycarp should increase the first number on 1, decrease the second number on 1, increase the third number on 1, and the fourth number should left unchanged. So, after Polycarp changed three elements by one, his sequence became equals to [25, 20, 15, 10], which is an arithmetic progression.

In the second example Polycarp should not change anything, because his sequence is an arithmetic progression.

In the third example it is impossible to make an arithmetic progression.

In the fourth example Polycarp should change only the first element, he should decrease it on one. After that his sequence will looks like [0, 3, 6, 9, 12], which is an arithmetic progression.

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

public class Codechef
{
   
    //static boolean vis[];
    int max = (int)1e5+1;
    static ArrayList<Integer>graph[];
    static int q_v[];
    static ArrayList<Integer>qs[];
    static int q_c[];
    static long arr[];
    static int temp[];
	public static void main (String[] args) throws java.lang.Exception
	{
	    InputReader input = new InputReader(System.in);
	    OutputWriter out = new OutputWriter(System.out);
	    int n = input.readInt();
	    int arr[] = new int[n];
	    for(int i =0;i<n;i++)arr[i] = input.readInt();
	    if( n<=2){System.out.println(0);return;}
	    int count =0;
	    boolean ok = false;
	    int res = Integer.MAX_VALUE;
	    for(int i=-1;i<2;i++){
	        int first = arr[0]+i;
	        for(int j=-1;j<2;j++){
	            count =0;
	            int sec = arr[1]+j;
	            count = sec==arr[1]?0:1;
	            count+= first == arr[0]?0:1;
	            int d = sec - first;
	            int z =0;
	            for( z =2;z<n;z++){
	                if(Math.abs(first+(z*d)-arr[z])<=1){
	                    count+=Math.abs(first+(z*d)-arr[z]);
	                }else break;
	            }
	           if(z == n){
	               res = Math.min(res,count);
	               ok = true;
	           }
	        }
	    }
	    if(ok)System.out.println(res);
	    else System.out.println(-1);
	}
	
	
}
 class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		
		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}
		
		public int readInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public String readString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		
		public double readDouble() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') {
				if (c == 'e' || c == 'E') {
					return res * Math.pow(10, readInt());
				}
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') {
				c = read();
				double m = 1;
				while (!isSpaceChar(c)) {
					if (c == 'e' || c == 'E') {
						return res * Math.pow(10, readInt());
					}
					if (c < '0' || c > '9') {
						throw new InputMismatchException();
					}
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
		
		public long readLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public boolean isSpaceChar(int c) {
			if (filter != null) {
				return filter.isSpaceChar(c);
			}
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		
		public String next() {
			return readString();
		}
		
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}
	
 class OutputWriter {
		private final PrintWriter writer;
		
		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}
		
		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}
		
		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0) {
					writer.print(' ');
				}
				writer.print(objects[i]);
			}
			writer.flush();
		}
		
		public void printLine(Object... objects) {
			print(objects);
			writer.println();
			writer.flush();
		}
		
		public void close() {
			writer.close();
		}
		
		public void flush() {
			writer.flush();
		}
	}
	   
//********************************************************************************************************************************
//////////////////////////////////NOTESSS//////////////////////////////////////////////////////////////////////
/*remember substring length you are dealing with take only those substrings
chain like pattern -> linked list (or) to from Arrays
finding degree can be done during input scanning itself(ok&=true:false flag trick)
longest subsquence can be solved in O(n) itself incase of consecutive numbers why try all subsequence go o dp[i-1]
(*******consecutive numberssssss*********)
initialize all global variables and collection ds

whenever deailing with AP remember first two terms define the entire series
*/
