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.
import sys
def gcd(a, b):
    while a > 1 and b > 1:
        a, b = max(a, b), min(a, b)
        a, b = b, a % b
    if a == 1 or b == 1:
        return 1
    if a == 0:
        return b
    if b == 0:
        return a

n = int(raw_input())
arr = map(int, raw_input().split())

g_gcd = reduce(gcd, arr)
if g_gcd > 1:
    print -1
    sys.exit(0)
one_count = arr.count(1)
if one_count > 0:
    print n - one_count
    sys.exit(0)

# possible to eventually get a 1
# but no 1s to start with
# ans is min # of steps to get a 1, plus n - 1

# claim: we have either:
# - two adjacent elements with gcd = 1
# -- OR, three adjacent elements a, b, c
# -- s.t. gcd(gcd(a,b),c) = 1, or gcd(a, gcd(b, c)) = 1
# for the global gcd to be 1, some new prime must be introduced at some point
# and, some prime must be "removed"
# 2 6 30 60 120 15
# 2, 2*3, 2*3*5, 2*2*3*5, 2*2*2*3*5, 3*5

best = n
for i in range(n):
    start = arr[i]
    for j in range(i, n):
        start = gcd(start, arr[j])
        if start == 1:
            best = min(best, j - i + 1)
            break
print best + n - 2

