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.
# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys

"""
created by shhuan at 2017/11/17 22:54

"""

N = int(input())
A = [int(x) for x in input().split()]
#
# N = 2000
# A = [random.randint(2, 5) for _ in range(N)]
#
# t0 = time.time()

def gcd(x, y):
    if x < y:
        x, y = y, x

    if y == 0:
        return x

    return gcd(x%y, y)

if any([v == 1 for v in A]):
    print(len([v for v in A if v != 1]))
else:

    d = float('inf')

    for i in range(N):
        v = A[i]
        for j in range(i+1, min(N, i+d+1)):
            v = gcd(v, A[j])
            if v == 1:
                d = min(d, j-i)

    if d != float('inf'):
        print(d + N - 1)
    else:
        print(-1)

    # print(time.time() - t0)


