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.
#	✪ H4WK3yE乡
#	Mayank Chaudhary
#	ABES EC , Ghaziabad
# ///==========Libraries, Constants and Functions=============///
import sys
from bisect import bisect_left,bisect_right
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log
from itertools import permutations
inf = float("inf")
mod = 1000000007
mini=1000000007

def fact(n):                                                             # <------  To calculate factorial of n under modulo m
    if n==0:
        return 1
    p=1;d=(10**9)+7
    for i in range(1,n+1):
        p=p*i
        p=p%d
    return p

def ncr(n,r):                                                          # < ------ To calculate nCr mod p value using Fermat Little under modulo m
    d=10**9+7
    num=fact(n)
    den=(fact(r)*fact(n-r))%d
    den=pow(den,d-2,d)
    return (num*den)%d

def sieve(n):                                                          # <----- sieve of eratosthenes for prime no.
    prime=[True for i in range(n+1)]
    lst=[0]*(n+1)
    p=2
    while p*p<=n:
        if prime[p]:
            lst[p]=p
            for i in range(p*p,n+1,p):
                if prime[i]==True:
                    prime[i]=False
                    lst[i]=p
        p=p+1
    for i in range(1,n+1):
        if lst[i]==0:
            lst[i]=i
    mylist=[0]*(n+1)
    for i in range(2,n+1):
        mylist[lst[i]]=mylist[lst[i]]+1
    return mylist
        
def binary(number):                                             #  <----- calculate the no. of 1's in binary representation of number
    result=0
    while number:
        result=result+1
        number=number&(number-1)
    return result

def calculate_factors(n):                                #<---- most efficient method to calculate no. of factors of number             
    hh = [1] * (n + 1); 
    p = 2; 
    while((p * p) < n): 
        if (hh[p] == 1): 
            for i in range((p * 2), n, p): 
                hh[i] = 0; 
        p += 1;  
    total = 1; 
    for p in range(2, n + 1): 
        if (hh[p] == 1):  
            count = 0; 
            if (n % p == 0): 
                while (n % p == 0): 
                    n = int(n / p); 
                    count += 1; 
                total *= (count + 1); 
    return total;

def prime_factors(n):                                         #<------------ to find prime factors of a no.
    i = 2
    factors = set()
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.add(i)
    if n > 1:
        factors.add(n)
    if len(factors)==2:
        return True
    else:
        return False

def isPrime(n):                                                      #<-----------check whether a no. is prime or not
    if n==2 or n==3: return True
    if n%2==0 or n<2: return False
    for i in range(3,int(n**0.5)+1,2):   # only odd numbers
        if n%i==0:
            return False    

    return True
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///==========MAIN=============///
n = int(input())
a = get_array()
if n < 3:
    print(0)
    exit()
r = []
for x in [-1, 0, 1]:
    for y in [-1, 0, 1]:
        f = 1
        t = 0
        s = a[0] + x
        d = a[1] + y - s
        b = range(s, s + n * d, d) if d else [s] * n
        for i in range(n):
            p = abs(b[i] - a[i])
            if p > 1:
                f = 0
                break
            if p == 1:
                t += 1
        if f:
            r.append(t)
print(min(r) if r else -1)