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 sys import setrecursionlimit
from collections import  Counter
setrecursionlimit(1000000)
d={}
def nod(a,b):
    if a==0:
        return b
    if b==0:
        return a
    if a==b:
        return a
    if a==1 or b==1:
        return 1
    if a%2==0:
        if b%2==0:
            return 2*nod(a//2,b//2)
        else:
            return nod(a//2,b)
    else:
        if b%2==0:
            return nod(b//2,a)
        else:
            if a>b:
                return nod(b,(a-b)//2)
            else:
                return nod((b-a)//2,a)
def nod2(a,b):
    try:
        return d[(a,b)]
    except:
        c=nod2(a,b)
        d[(a,b)]=c
        return c
n=int(input())
a=list(map(int,input().split()))
z=a[0]
for i in a:
    z=nod(z,i)
if z!=1:
    print(-1)
else:
    z=0
    if 1 in a:
        print(n-Counter(a)[1])
        exit()
    else:
        while not 1 in a:
            for i in range(len(a)-1):
                a[i]=nod(a[i],a[i+1])
            a.pop()
            z=z+1
        print(n+z-1)
