You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say x and y, and replace one of them with gcd(x, y), where gcd denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples

Input

5
2 2 3 4 6


Output

5


Input

4
2 4 6 8


Output

-1


Input

3
2 6 9


Output

4

Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

  * [2, 2, 3, 4, 6]. 
  * [2, 1, 3, 4, 6]
  * [2, 1, 3, 1, 6]
  * [2, 1, 1, 1, 6]
  * [1, 1, 1, 1, 6]
  * [1, 1, 1, 1, 1]



We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
'''
You have an array of integers
You need to perform operation on these numbers so that all becomes 1
operation: replace any two adjacent numbers with GCD of these two numbers
print number of operation needed or -1 if it can not be done
Examples
input
5
2 2 3 4 6
output
5
input
4
2 4 6 8
output
-1
input
3
2 6 9
output
4
startegy: GCD of any two numbers would be 1 if either of them is 1
or if all of them are primes
If even one 1 is avaible in array , it can be used to convert others to one
In that case nop needed is number of element -1
So first count how many 1 is available . Answer would be len array - count
In case no 1 is available startegy would be to sequentially try to convert
number with their gcd till we get 1 as gcd.
nop = nop done + length of remaining array
'''


from math import gcd
n=int(input())
a=list(map(int, input().split()))
onecnt = a.count(1)
if onecnt > 0:
    print(n - onecnt)
else:
    nop = 0
    flag=True
    while (a.count(1)==0):
        if (len(a)-nop==1 and a[0]!=1):
            flag = False
            break
        nop+=1
        for i in range(len(a)-nop):
            a[i]=gcd(a[i], a[i+1])
    print([-1,nop+n -1][flag])
