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.
#trying to find the samllest segment whose gcd is 1
#not gonna use segment tree instead we  will use a faster method to find gcd which includes prefix and suffix of gcd
#refer to seg_gcd2.py in nov17

from math import gcd

def check(val):
	for i in range(0,n,val):
		l=i
		r=min(n-1,i+val-1)
		pre[l]=arr[l]
		for j in range(l+1,r+1):
			pre[j]=gcd(pre[j-1],arr[j])
		suf[r]=arr[r]
		for j in range(r-1,l-1,-1):
			suf[j]=gcd(suf[j+1],arr[j])
	i=0
	#print(val,n)
	while True:
		if i+val-1>=n:
			break
		if gcd(suf[i],pre[i+val-1])==1:
			return True
		i+=1
	return False	


def binS():
	l=0
	r=n+1
	flag=False
	while l < r:
		mid=(l+r)>>1
		if mid < 1:
			break
		if check(mid):
			flag=True
			r=mid
		else:
			l=mid+1
	if flag:	
		return r
	else:
		return -1
n=int(input())
arr=[int(x) for x in input().split()]
k=0
for i in arr:
	if i==1:
		k+=1
pre=[None]*n
suf=[None]*n
ans=binS()
if ans==1:
	print(n-k)
elif ans ==-1:
	print(-1)
else:
	print(ans-1+n-1)
