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(''))
s=list(map(int,input('').split()))
a=s.count(1)
ans=0
def gcd(x,y):
    x,y=max(x,y),min(x,y)
    if y==0:
        return x
    return gcd(y,x%y)
if a!=0:
    ans=n-a
else:
    dp=[[0 for i in range(n)] for j in range(n)]
    for i in range(n):
        dp[i][i]=s[i]
    mins=n
    for i in range(n):
        for j in range(i,n):
            dp[i][j]=gcd(dp[i][j-1],s[j])
            if dp[i][j]==1:
                mins=min(j-i,mins)
    if mins==n:
        ans=-1
    else:
        ans=mins+(n-1)
print(ans)
    
    
