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.
from fractions import gcd

n = int(raw_input(''))
tab_a = map(int, raw_input('').split(' '))
przed = 999999
num_1 = 0
for x in range(0, len(tab_a), +1):
    if tab_a[x] == 1:
        num_1 += 1
if num_1 != 0:
    print n-num_1
    exit(0)
for x in range(0, len(tab_a)-1, +1):
    c = gcd(tab_a[x], tab_a[x+1])
    if c == 1:
        przed = 1
    for y in range(x+2, len(tab_a), +1):
        c = gcd(c, tab_a[y])
        if c == 1:
            if y-x <= przed:
                przed = y-x
        #print x,y
if przed == 999999:
    print '-1'
else:
    print (przed)+(n-1)