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;
int a[2010];
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int main() {
  int n;
  while (scanf("%d", &n) > 0) {
    int not_one = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      not_one += (a[i] != 1);
    }
    if (!not_one) {
      printf("0\n");
      continue;
    }
    if (not_one < n) {
      printf("%d\n", not_one);
      continue;
    }
    int min_seg = n + 1;
    for (int i = 0; i < n; i++) {
      int tmp = a[i], j = i + 1;
      while (j < n and tmp != 1) {
        tmp = gcd(tmp, a[j]);
        j++;
      }
      if (tmp == 1) {
        min_seg = min(min_seg, j - i);
      }
    }
    if (min_seg == n + 1) {
      printf("-1\n");
    } else {
      printf("%d\n", not_one + min_seg - 2);
    }
  }
}
