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 gcd(int a, int b) { return (a % b == 0) ? b : gcd(b, a % b); }
int main(void) {
  std::ios::sync_with_stdio(false);
  int arr[2020];
  int n, totone = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) {
      totone++;
    }
  }
  int mini = 1000000000;
  if (totone) {
    cout << n - totone << endl;
  } else {
    for (int i = 0; i < n; i++) {
      int g = arr[i];
      for (int j = i; j < n; j++) {
        g = gcd(g, arr[j]);
        if (g == 1) {
          mini = min(mini, n - 1 + j - i);
          break;
        }
      }
      if (!i && g != 1) {
        cout << -1 << endl;
        return 0;
      }
    }
    cout << mini << endl;
  }
  return 0;
}
