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: int, b: int): return gcd(b, a % b) if b else a

nb = int(input())
s = input()
numbers = [int(i) for i in s.split(" ")]
number = -1

flag = True

op = 0

def count1(s):
    res = 0
    for i in s:
        if i == 1:
            res += 1
    return res

if nb == 1:
    if numbers[0] == 1:
        print(0)
    else:
        print(-1)
else:
    while op < 2010:
        new = []
        for i in range(0, len(numbers) - 1):
            g = gcd(numbers[i], numbers[i + 1])
            if g == 1:
                number = nb + op - count1(numbers)
                numbers = []
                print(number)
                exit(0)
            else:
                new.append(g)

        numbers = new
        op += 1

    print(number)