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
import math
import sys
n = int(input())
a = input().split()
m = list()
pr = 0
hi = False
for i in a:
    ia = int(i)
    m.append(ia)
    if not hi and ia % 2 == 0:
        pr += 1
    else:
        hi = True
if pr == n:  # Debe haber al menor un par.
    print(-1)
    sys.exit()
m2 = m
ps = 0


def process():
    pk = 0
    for j in range(n - 1):  # Primero revisa únicamente unos.
        a1, a2 = m[j], m[j + 1]
        gd = math.gcd(a1, a2)
        if gd == 1:
            return n
        else:
            if a1 % 2 != 0:  # CASO 1: el primero es impar.
                if a2 % 2 == 0:  # CASO 1a: el segundo es par.
                    m2[j + 1] = gd
                else:  # CASO 1b: ambos son impares. Busca mayor.
                    if a1 == max(a1, a2):
                        m2[j] = gd
                    else:
                        m2[j + 1] = gd
                pk += 1
            elif a2 % 2 != 0:  # CASO 2: el segundo es impar.
                m2[j] = gd
                pk += 1
            a1, a2 = m[j], m[j + 1]  # Vuelve a asignar.
            gd = math.gcd(a1, a2)  # Vuelve a revisar.
            if gd == 1:
                return n + pk
    return pk


while True:
    rs = process()
    if rs == n:
        print(str(n + ps))
        break
    elif rs > n:
        print(str(rs + ps))
    else:
        ps += rs
        m = m2
