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>
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const double eps = 1e-18;
const double PI = acos(-1.0);
using namespace std;
int a[N], f[N];
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int d = mod, g = 0;
  int aa = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 1) aa++;
  if (aa > 0) return 0 * printf("%d\n", n - aa);
  for (int i = 0; i < n; i++) g = gcd(g, a[i]);
  if (g > 1) return 0 * puts("-1");
  for (int i = 0; i < n; i++) {
    g = a[i];
    for (int j = i + 1; j < n; j++) {
      g = gcd(g, a[j]);
      if (g == 1) d = min(d, j - i);
    }
  }
  if (d == n + 1)
    return 0 * puts("-1");
  else
    printf("%d\n", d + n - 1);
  return 0;
}
