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(raw_input())
a=[int(k) for k in raw_input().split(" ")]


def pgcd(a,b) :              # retourne g = pgcd(a,b)
    if b == 0 : return a#(a,1,0)
    #q = a // b 
    r = a % b
    g=pgcd(b,r)#(g,u1,v1) = bezout(b,r)    
    return g#(g,v1,u1-q*v1) 


if min(a)==1:
    print n-len([k for k in a if k==1])
else:
    bests=[0]*n
    bb=n+1
    left={}
    for i in range(n):
        left[i]=a[i]
        for k in left.keys():
            bests[k]+=1
            left[k]=pgcd(a[i],left[k])
            if left[k]==1:
                bb=min(bb,bests[k])
                del left[k]
            elif bests[k]>=bb:
                del left[k]
    
    #print bests
    #print left
    if bb==n+1:
        print -1
    else:
        print n-2 + min([bests[k] for k in range(n) if k not in left])