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.
#Complexity - O(n^2)
from collections import Counter
import math

n = int(input())
arr = list(map(int, input().split()))
counter = Counter(arr)
if counter[1] > 0:
  print(n- counter[1])
  exit(0)

def find_result(arr):
  counter = 1
  while len(arr) > 1:
    newarr = []
    for i in range(len(arr)-1):
      hcf = math.gcd(arr[i], arr[i+1])
      if hcf == 1:
        return counter
      newarr.append(hcf)
    counter += 1
    arr = newarr
  return -1 if arr[0] != 1 else counter

rounds_to_achieve_1 = find_result(arr)
if rounds_to_achieve_1 == -1:
  print(-1)
else:
  print(len(arr)-1+rounds_to_achieve_1)
