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?
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 *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>) — the elements of the array.
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.
5
2 2 3 4 6
5
4
2 4 6 8
-1
3
2 6 9
4
15
10 10 10 10 10 10 21 21 21 21 21 21 21 21 21
15
12
10 10 14 14 14 14 14 14 14 14 21 21
20
5
10 10 14 21 21
6
9
10 10 10 10 10 14 14 21 21
11
9
10 10 10 10 10 10 10 10 21
9
13
10 10 10 15 15 15 15 15 15 15 15 21 21
21
15
10 10 10 10 10 10 10 10 10 10 10 10 15 15 21
17
4
1 1 1 1
0
1
3
-1
2
1 1
0
2
1000000000 1000000000
-1
1
1000000000
-1
1
1
0
3
42 15 35
4
3
6 10 15
4
4
2 1 1 1
1
5
2 1 1 1 2
2
3
30 14 21
4
3
15 6 10
4
4
1 1 1 2
1
5
1 1 1 2 2
2
4
2 6 9 1
3
6
2 3 4 1 1 1
3
15
2 6 6 6 3 3 3 15 5 5 5 7 5 5 5
15
5
2 3 2 6 9
5
6
6 15 10 6 15 10
7
def gcd(a,b):
	if b==0:
		return a
	else:
		return gcd(b,a%b)
N=int(input())
a=[int(x) for x in input().split()]
for i in range(N-1):
	if gcd(a[i],a[i+1])==1:
		print(N)
		exit()
for i in range(N-2):
	if gcd(gcd(a[i],a[i+1]),a[i+2])==1:
		print(N+1)
		exit()
print("-1")