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;
long long int arr[100001];
inline long long int gcd(long long int a, long long int b) {
  long long int tmp;
  while (b) {
    tmp = a % b;
    a = b;
    b = tmp;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(false);
  ;
  cin.tie(0);
  ;
  long long int i, j, n, x, y, cnt = 0, mini;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
    cnt += (arr[i] == 1);
  }
  if (cnt > 0) {
    cout << n - cnt;
    return 0;
  }
  mini = n + 1;
  for (i = 1; i <= n; i++) {
    x = arr[i];
    for (j = i + 1; j <= n; j++) {
      x = gcd(x, arr[j]);
      if (x == 1) {
        mini = min(mini, j - i + 1);
        break;
      }
    }
  }
  if (mini == n + 1) {
    cout << "-1";
    return 0;
  }
  cout << mini - 1 + n - 1;
  return 0;
}
