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 = [int(i) for i in input().split()]
least = len(a)
found = 0
count = 0
d={}
def gcd(x,y):
	if x<y:
		x,y=y,x
	if (x,y) not in d:
		if x%y==0:
			d[(x,y)]=y
		else:
			#print(x,y)
			d[(x,y)]=gcd(y,x%y)
	return d[(x,y)]

for i in range(len(a)):
	l=a[i]
	if a[i]==1:
		count+=1
	for j in range(i,min(len(a),least+i)):
		l = gcd(a[j],l)
		if l==1:
			found = 1
			least = min(least,j-i)

#print(d,least)
if found==0:
	print(-1)	
elif count>0:
	print(n-count)
else:
	print(least+n-1)