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 ) )

memo = {}

def is_prime(n):

    if n < 1:
        return False

    if n == 1:
        return ( True )

    if n in memo:
        return memo[n]

    i = 2

    while i*i <= n:
        if n % i == 0:
            memo[n] = False
            return False
        i += 1

    memo[n] = True

    return True

n = int( input() )
a = list( map( int, input().split() ) )

tot = 0
for i in range( n ):
    if a[i] == 1:
        tot += 1

if tot > 0:
    print( n-tot )
    exit(0)

tot = 0

best = 10**9

for i in range( n ):
    cur = a[i]
    for j in range( i+1, n ):
        cur = gcd( cur, a[j] )
        if cur == 1:
            best = min( best,j-i )

if best == 10**9:
    print( "-1" )
else:
    print( n+best-1 )
    
