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.
def gcd(a,b):
    if b==0:
        return a
    else:
        return gcd(b,a%b)
    
n = int(raw_input())
l1 = map(int, raw_input().split())
l = []
for i in l1:
    if i != 1:
        l.append(i)
ans = 0
if len(l1) > len(l):
    print len(l)
else:
    if len(l)==1 and l[0]!= 1:
        print -1
    else:
        ans = 10000
        for i in range(n):
            temp = l[i]
            for j in range(i+1,min(n,i+ans+1)):
                temp = gcd(temp,l[j])
                if temp == 1:
                    ans = min(ans,j-i)
                    break
        if ans == 10000:
            print -1
        else:
            print ans + n - 1