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())
A = list(map(int, input().split()))
if n == 1:
    if A[0] != 1:
        print(-1)
    else:
        print(0)
    exit()

import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)

def gcd_list(numbers):
    return reduce(math.gcd, numbers)

g = gcd_list(A)
if g != 1:
    print(-1)
    exit()

cnt = 0
for i in range(n):
    if A[i] == 1:
        cnt += 1
if cnt:
    print(n-cnt)
    exit()

while len(A) >= 2:
    B = []
    for i in range(len(A)-1):
        B.append(gcd(A[i], A[i+1]))
    for i in range(len(B)):
        if B[i] == 1:
            print(n-len(B)+n-1)
            exit()
    A = B
