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.
def gcd(a, b):
	if b == 0:
		return a
	return gcd(b , a%b)

nums = int(input())
array = input().split()
flag = False 
cnt = 0 
for obj in array:
	if int(obj) == 1:
		flag = True
	else:
		cnt += 1

if cnt != len(array):
	print(cnt)
else :
	minn_time = 0x7fffffff
	for i in range(0 , len(array)):
		if i != len(array) - 1:
			p = int(array[i + 1])
		else :
			p = int(array[i])
		for j in range(i + 1 , len(array)):
	#		if i == len(array) - 1:
	#			print("we are here ")
			if j == i + 1:
				p = gcd(p , int(array[i]))
			else :
				p = gcd(p , int(array[j]))
			if p == 1:
			#	print("we are here")
				minn_time = min(minn_time , j - i)
				flag = True
			#	break
	if minn_time == 0x7fffffff :
		print("-1")
	else :
		print(minn_time + cnt - 1)