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 num1, int num2) {
  if (num2 == 0) return num1;
  return gcd(num2, num1 % num2);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int _gcd = a[0];
  for (int i = 1; i < n; i++) {
    _gcd = gcd(_gcd, a[i]);
  }
  if (_gcd != 1) {
    cout << -1 << endl;
    return 0;
  }
  int num_ones = 0;
  for (int x : a) {
    if (x == 1) num_ones++;
  }
  if (num_ones >= 1) {
    cout << n - num_ones << endl;
    return 0;
  }
  int min_steps = INT_MAX;
  for (int start = 0; start < (n - 1); start++) {
    int ansr = a[start], num_steps = 0;
    for (int stop = start + 1; stop < n; stop++) {
      num_steps++;
      ansr = gcd(ansr, a[stop]);
      if (ansr == 1) break;
    }
    if (ansr == 1) min_steps = min(min_steps, num_steps);
  }
  cout << min_steps + n - 1 << endl;
  return 0;
}
