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.
a = int(input())
s = list(map(int, input().split()))
def g(a, b):
    a, b = max(a, b), min(a, b)
    while a % b > 0:
        a, b = b, a % b
    return b
b = 0
c = 0
for i in range(1, a):
    if g(s[i], s[i - 1]) == 1:
        b = 1
        break
if s.count(1) > 0 or b:
    b = 1
else:
    for j in range(1, a):
        for i in range(1, a):
            v = g(s[i], s[i - 1])
            if s[i] != s[i - 1]:
                if s[i] == 1:
                    s[i - 1] = v
                else:
                    s[i] = v
                c += 1
            if v == 1:
                b = 1
                break
        if s[-1] == 1:
            b += 1
        if b:
            break
if b or a == 1 and s[0] == 1:
    print(c + a - s.count(1))
else:
    print(-1)