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 max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int power(long long int n, long long int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long int res = power(n, k / 2);
  if (k % 2 == 0) return res * res;
  return res * res * n;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int mini = INT_MAX;
  int o = 0;
  for (int i = 0; i < n; i++)
    if (arr[i] == 1) o++;
  if (o > 0) {
    cout << n - o << endl;
    return 0;
  }
  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, j - i);
        break;
      }
    }
  }
  if (mini == INT_MAX)
    cout << -1;
  else
    cout << n - 1 + mini;
  return 0;
}
