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.
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int mod = 1e9 + 7;
int x, n, ans, len, a[N];
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++) scanf("%d", &a[i]);
  len = n + 1;
  for (int i = (1); i < (n + 1); i++) {
    x = a[i];
    for (int j = (i); j < (n + 1); j++) {
      x = gcd(x, a[j]);
      if (x == 1) {
        len = min(len, j - i);
        break;
      }
    }
  }
  for (int i = (1); i < (n + 1); i++)
    if (a[i] == 1) ans--;
  if (len == 0)
    ans = ans + n;
  else
    ans = ans + n + len - 1;
  if (len == n + 1)
    printf("-1");
  else
    printf("%d", ans);
  return 0;
}
