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.
import math

def iterArr(arr, num):
    dist = num
    for i in range(num):
        for j in range(i+1, num):
            if math.gcd(arr[i], arr[j]) == 1:
                if j-i < dist:
                    dist = j-i
    return dist

n = int(input())
arrIn = list(map(int, input().split()))
i = arrIn.count(1)
if i > 0:
    print(n-i)
else:
    dist1 = iterArr(arrIn, n)
    if dist1 == n:
        arr1 = []
        for k in range(n-1):
            arr1.append(math.gcd(arrIn[k], arrIn[k+1]))
        lenArr1 = len(arr1)
        dist2 = iterArr(arr1, lenArr1)
        if dist2 == lenArr1:
            print(-1)
        else:
            print(dist2+n)
    else:
        print(n-1+dist1)