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.
n = int(input())
A = input().split()
A = [int(i) for i in A]
def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a % b)
C = [n+10000]*n
for i in range (0,n):
    B = A[:]
    j = i
    while B[i] != 1:
        j+=1
        if j >= n:
            break
        else:
            B[i] = gcd(B[i],A[j])
    if B[i] == 1: C[i] = j-i
    else: C[i] = n+100
small = min(C)
if small == n+100:
    print (-1)
elif small ==0:
    print (n-C.count(0))
else:
    print (n-1+small)