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 MAXN = 2005;
int n, a[MAXN], ans = 1e9, C;
int GCD(int aa, int bb) { return !aa ? bb : GCD(bb % aa, aa); }
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], C += (a[i] == 1);
  for (int i = 0; i < n; i++) {
    int gc = a[i];
    for (int j = i; j < n; j++) {
      gc = GCD(gc, a[j]);
      if (gc == 1) {
        ans = min(ans, j - i);
        break;
      }
    }
  }
  if (ans == 1e9)
    cout << -1;
  else
    cout << ans + n - (C + !(ans == 0));
}
