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 math
import sys
from bisect import bisect_right, bisect_left, insort_right
from collections import Counter, defaultdict
from heapq import heappop, heappush
from itertools import accumulate
from sys import stdout

R = lambda: map(int, input().split())
n = int(input())
arr = list(R())
md = -1
ones = arr.count(1)
if ones:
    print(n - ones)
    exit(0)
dp = [[0 for j in range(n)] for i in range(n)]
ml = n + 1
for l in range(1, n + 1):
    for i in range(n - l + 1):
        if l == 1:
            dp[i][i] = arr[i]
        else:
            dp[i][i + l - 1] = min(math.gcd(dp[i][i + l - 2], arr[i + l - 1]), math.gcd(dp[i + 1][i + l - 1], arr[i]))
        if dp[i][i + l - 1] == 1:
            ml = min(ml, l)
            break
if ml == n + 1:
    print(-1)
else:
    print(n + ml - 2)